problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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2. Solve the system $\left\{\begin{array}{c}\sin x \cdot \sin (x y)=-1 \\ 6 y^{2}=y \operatorname{tg}(x / 2)+1\end{array}\right.$. | Answer: $\left\{\begin{array}{c}x=\frac{3 \pi}{2}(4 t+1), \\ y=\frac{1}{3},\end{array} t \in Z\right.$. | {\begin{pmatrix}\frac{3\pi}{2}(4+1),\\\frac{1}{3},\end{pmatrix}\inZ.} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,372 |
3. It is known that for some positive coprime numbers $m$ and $n$, the numbers $m+1947 n$ and $n+1947 m$ have a common prime divisor $d>9$. Find the smallest possible value of the number $d$ under these conditions. | Answer: $d_{\min }=139$.
For example,
$$
\begin{aligned}
& m=138, n=1 \rightarrow 1947 m+n=1947 \cdot 138+1=139 \cdot 1933 \\
& m+1947 n=138+1947=139 \cdot 15
\end{aligned}
$$ | 139 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,373 |
5. For what values of $a$ does the equation
$$
x^{3}+48(a+2) x+32(a-1)=0
$$
have exactly two roots? Find these roots. | Answer: 1) $a=-3$; 2) $x_{1}=-4, x_{2}=8$. | =-3;x_{1}=-4,x_{2}=8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,375 |
6. On the side $A C$ of triangle $A B C$, there is a point $D$ such that $A D: A C=2: 5$, and $B D+2 B C=3 A B$. The inscribed circle of the triangle with center at point $O$ intersects $B D$ at points $M$ and $N$. Find the angle $M O N$. | Answer: $2 \arccos \frac{1}{5}$.
Examples of tasks from the database of the remote qualifying round of the "Rosatom" Olympiad, 11th grade
The database of the remote qualifying round of the "Rosatom" Olympiad (which is only for 11th-grade students) contains more than 300 problems with numerical answers (which are the ... | 2\arccos\frac{1}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,376 |
5. The lengths of the sides of a parallelogram are 3 and 2. The bisectors of all its internal angles limit a polygon on the plane. Find the ratio of its area to the area of the parallelogram.
## Answers and solutions
Problem 1 Answer: in 16 min. | Solution $v_{A}, v_{k}, v_{c}$ - the speeds of the bus, Kolya, and Sasha respectively, $T$ - the required time.
The distance between Kolya and Sasha at the moment the bus stops: $2\left(v_{A}+v_{k}\right)$. The speed of their approach: $v_{C}-v_{k}, v_{C}=3 v_{k}, v_{A}=5 v_{C}=15 v_{k}-$ conditions of the problem. $T... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,379 |
Problem 6. Answer: $|L|_{\text {min }}=\frac{31}{2}=15.5$.
## Variant 0
The base $A B C$ of the pyramid $A B C D$ lies on the plane $S$. Point $M$ is located on the extension of the line $A B$ beyond point $B$ such that $M B: A B=\lambda$. Point $N$ is the midpoint of the edge $D C$. The curve $L$, connecting points ... | Answer: $|L|_{\min }=\frac{a}{2} \sqrt{4 \lambda^{2}+6 \lambda+3}$.
Solution. Let the curve $L$ be the desired one. It cannot fail to intersect the line $A B$ at some point $E$. Then the part of the curve from point $M$ to point $E$ is a straight line segment. Point $F$ is the midpoint of edge $A C$. The geometric loc... | \frac{}{2}\sqrt{4\lambda^{2}+6\lambda+3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,388 |
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 | 2,400 |
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 | 2,403 |
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 | 2,405 |
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 | 2,407 |
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 | 2,408 |
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 | 2,409 |
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 | 2,410 |
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 | 2,411 |
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 | 2,421 |
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 | 2,422 |
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 | 2,428 |
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 | 2,429 |
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 | 2,430 |
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 | 2,431 |
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 | 2,432 |
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 | 2,433 |
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 | 2,434 |
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 | 2,435 |
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
$$
\begin{aligned}
& k=\frac{7 n-3}{5}=n+\frac{2 n-3}{5} \rightarrow \frac{2 n-3}{5}=l \in Z \rightarrow n=\frac{5 l+3}{2}=2 l+1+\frac{l+1}{2} \rightarrow \\
& \rightarrow \frac{l+1}{2}=t ... | 5-1,\neq3-1,,\inN | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,436 |
4. Represent the number 43 as the sum of three prime numbers. In how many ways can this be done? Note that the number 1 is not considered a prime number. | Solution.
$$
43=a+b+c, a, b, c \text { - prime numbers. }
$$
Prime numbers from 2 to 43
| 2 | 3 | 5 | 7 | 11 | 13 | 17 | 19 | 23 | 29 | 31 | 37 | 41 | 43 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
$a=2 \rightarrow 43-4=41 \rightarrow b=2 \righ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,437 |
1. Petya came to the bus stop of the bus going to school with stops equally spaced from each other, and, not seeing the bus on the road, decided to run and get on the bus at the next stops along the way to school. Petya ran in such a way that at any moment he could notice the appearance of the bus on the road behind hi... | Answer: $a_{\max }=0.6$ km | 0.6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 2,440 |
2. The coordinates $(x ; y)$ of points in the square $\{(x ; y):-\pi \leq x \leq \pi, 0 \leq y \leq 2 \pi\}$ satisfy the system of equations $\left\{\begin{array}{c}\sin x+\sin y=\sin 2 \\ \cos x+\cos y=\cos 2\end{array}\right.$. How many such points are there in the square? Find the coordinates $(x ; y)$ of the point ... | Answer: 1) two points
$$
\text { 2) } x=2+\frac{\pi}{3}, y=2-\frac{\pi}{3}
$$ | 2)2+\frac{\pi}{3},2-\frac{\pi}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,441 |
6. Along the diagonals of the bases $AC$ and $B_{1}D_{1}$ of the cube $ABCD A_{1}B_{1}C_{1}D_{1}$ with edge $a$, two ants, Gosha and Lesha, are crawling. They started moving simultaneously from points $A$ and $B_{1}$ respectively at a constant speed, with Lesha's speed being three times the speed of Gosha's movement, a... | Answer: $d_{\min }=a \sqrt{\frac{6}{5}}$ | \sqrt{\frac{6}{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,443 |
1. Petya came to the bus stop of the bus going to school with stops equally spaced from each other, and, not seeing the bus on the road, decided to run and get on the bus at the next stops along the way to school. Petya ran in such a way that at any moment he could notice the appearance of the bus on the road behind hi... | Answer: $a_{\max }=0.25$ km | 0.25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 2,444 |
2. The coordinates $(x ; y)$ of points in the square $\{(x ; y):-\pi \leq x \leq 2 \pi, 0 \leq y \leq 3 \pi\}$ satisfy the system of equations $\left\{\begin{array}{l}\sin x+\sin y=\sin 3 \\ \cos x+\cos y=\cos 3\end{array}\right.$. Find the coordinates $(x ; y)$ of all such points. | Answer: 1) five points;
$$
\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 | 2,445 |
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 | 2,446 |
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 | 2,447 |
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 | 2,448 |
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 | 2,449 |
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 | 2,451 |
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 | 2,456 |
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 | 2,457 |
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 | 2,459 |
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 | 2,460 |
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 | 2,463 |
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 | 2,464 |
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 | 2,465 |
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 | 2,466 |
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 | 2,467 |
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 | 2,468 |
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 | 2,469 |
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 | 2,470 |
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 | 2,472 |
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 | 2,473 |
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 | 2,476 |
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 | 2,477 |
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 | 2,480 |
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
$$
\sin \left(x-\frac{\pi}{4}\ri... | Answer: 1) $x=\frac{\pi}{2}+2 \pi k, k=0,1,2,3$; 2) $x=\pi(2 m+1), m \in Z$. | \frac{\pi}{2}+2\pik,k=0,1,2,3;\pi(2+1),\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,481 |
4. Let's calculate the number of ways to write 3 digits:
$$
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 | 2,482 |
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
$$
\cos (x+y)+\sin (x-y) \cos (x+y)=1+\s... | Answer: $(x+y)_{\min }=2 \pi$. | (x+y)_{\}=2\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,483 |
6. Let the length of side $AB$ of parallelogram $ABCD$ be denoted by $a$. Drop a perpendicular $BH$ from vertex $B$ to line $CM$. Consider the right triangle $BHC$. According to the problem, $BH = AB = a$. Since $BC \parallel AB$, we have $\angle BHC = \angle CMD = 30^{\circ}$. Therefore, $BC = 2BH = 2a$. Now consider ... | Answer:
1) $60^{\circ}, 120^{\circ}$; 2) $S_{A B C D}=\sqrt{3}$
# | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,484 |
2. Solve the equation
$$
\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 | 2,485 |
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 | 2,486 |
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 | 2,488 |
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 | 2,490 |
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 | 2,491 |
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 | 2,494 |
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 | 2,496 |
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 | 2,497 |
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 | 2,498 |
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 | 2,499 |
2. For what values of $a$ is the point with coordinates $(\sin 2 a ; \cos 3 a)$ symmetric to the point with coordinates $(\sin 3 a ; \cos 2 a)$ with respect to the y-axis. | Answer: $a=\frac{2 \pi n}{5}, n \in Z$. | \frac{2\pin}{5},n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,501 |
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=1: 3, A Q: Q C=1: 1$. 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 six times. Find the expected va... | Answer: 1) $\left.P(A)=\frac{5}{6} ; 2\right) M_{X}=\frac{1}{4}$. | P(A)=\frac{5}{6};M_{X}=\frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,502 |
5. Represent the number 1947 as the sum of the cubes of five integers. Prove that any integer can be represented as the sum of the cubes of five integers. | Answer: $1947=(321)^{3}+(319)^{3}+(-320)^{3}+(-320)^{3}+(3)^{3}$. | 1947=(321)^{3}+(319)^{3}+(-320)^{3}+(-320)^{3}+(3)^{3} | Number Theory | proof | Yes | Yes | olympiads | false | 2,503 |
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=1: 4, A Q: Q C=3: 1$. 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 two times. Find the expected va... | Answer: 1) $\left.P(A)=\frac{2}{11} ; 2\right) M_{X}=\frac{13}{40}$. | P(A)=\frac{2}{11};M_{X}=\frac{13}{40} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,505 |
5. Point $M$ divides side $B C$ of parallelogram $A B C D$ in the ratio $B M: M C=2$. Line $A M$ intersects diagonal $B D$ at point $K$. Find the area of quadrilateral $C M K D$, if the area of parallelogram $A B C D$ is 1.
Task 1 Answer: 236 minutes or 173 minutes. | Solution. It was considered correct to count the moment 00:00 as both happy and not happy, with the solution practically unchanged.
Happy moments when the number of minutes is three times the number of hours: 01:03, $02:06, \ldots, 19:57$ and, perhaps, 00:00. The intervals between these happy moments are 1 hour 3 minu... | \frac{11}{30} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,506 |
2. Petya wrote down five consecutive terms of an arithmetic progression and encrypted them according to the principle: each digit was replaced with a letter, different digits with different letters, and vice versa. Here's what he got: A, BS, DE, AV, CB. What numbers did Petya write? | Answer: $3,15,27,39,51$.
Solution. From the condition, it follows that the terms of the arithmetic sequence are natural numbers and the difference of the progression, denoted as $d$, is positive. We will consider the letters A, B, C, D, E as digits or numbers, and the corresponding two-digit numbers will be denoted as... | 3,15,27,39,51 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 2,508 |
5. Point $M$ divides side $B C$ of parallelogram $A B C D$ in the ratio $B M: M C=1: 2$. Line $A M$ intersects diagonal $B D$ at point $K$. Find the area of quadrilateral $C M K D$, if the area of parallelogram $A B C D$ is 1. | Answer: $\frac{11}{24}$.
# | \frac{11}{24} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,509 |
2. Petya wrote down five consecutive terms of an arithmetic progression and encrypted them according to the principle: each digit was replaced with a letter, different digits with different letters, and vice versa. Here's what he got: A, BV, DS, EE, SD. What numbers did Petya write? | Answer: $6,15,24,33,42$. | 6,15,24,33,42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 2,510 |
5. Point $M$ divides side $B C$ of parallelogram $A B C D$ in the ratio $B M: M C=3$. Line $A M$ intersects diagonal $B D$ at point $K$. Find the area of quadrilateral $C M K D$, if the area of parallelogram $A B C D$ is 1. | Answer: $\frac{19}{56}$.
# | \frac{19}{56} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,511 |
2. Petya wrote down five consecutive terms of an arithmetic progression and encrypted them according to the principle: each digit was replaced with a letter, different digits with different letters, and vice versa. Here's what he got: A, BV, BS, VD, EE. What numbers did Petya write? | Answer: $5,12,19,26,33$.
Solution. From the condition, it follows that the terms of the arithmetic sequence are natural numbers and the difference of the progression, denoted as $d$, is positive. We consider the letters A, B, V, C, D, E as digits or numbers, and the corresponding two-digit numbers will be denoted as $... | 5,12,19,26,33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 2,513 |
5. Point $M$ divides side $B C$ of parallelogram $A B C D$ in the ratio $B M: M C=1: 3$. Line $A M$ intersects diagonal $B D$ at point $K$. Find the area of quadrilateral $C M K D$, if the area of parallelogram $A B C D$ is 1. | Answer: $\frac{19}{40}$.
## Grading Criteria for the Final Round of the Rosatom Olympiad 04.03.2023, 9th Grade
In all problems, the correct answer without justification $\quad 0$ | \frac{19}{40} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,514 |
1. $5|\cos 2 x|$ from natural values can only take the values $1,2,3,4,5$. Since
$\text{GCD}(5|\cos 2 x|, a|\sin 2 x|)=3$
then $5|\cos 2 x|=3$, i.e.
$|\cos 2 x|=3 / 5 \Rightarrow|\sin 2 x|=4 / 5$ and $x= \pm \frac{1}{2} \arccos \frac{3}{5}+\frac{\pi n}{2}, n \in Z$.
For the parameter $a$ we get the equation
$$
\te... | Answer: $\left\{\begin{array}{l}a=15 k, k \in N ; \\ x= \pm \arccos \frac{3}{5}+\frac{\pi n}{2}, n \in Z .\end{array}\right.$ | {\begin{pmatrix}=15k,k\inN;\\\\arccos\frac{3}{5}+\frac{\pin}{2},n\inZ0\end{pmatrix}.} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,515 |
3. Since the parabola intersects the x-axis at $x=1$, the value of the function $y=a x^{2}+b x+c$ at this point is zero, i.e.,
$$
a+b+c=0
$$
The equation of the given line can be written as
$$
y=5-2 x
$$
The condition for the line to be tangent to the parabola at the point $(x ; y)$ is that the values of the functi... | Answer: $\left(-\frac{3}{5} ; \frac{16}{3}\right),\left(\frac{5}{3} ; \frac{4}{3}\right)$. | (-\frac{1}{3};\frac{16}{3}),(\frac{5}{3};\frac{4}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,517 |
5. The first equation of the system defines a pair of parallel lines when $\mathrm{b} \neq 0$ and one line when $b=0$. The second equation defines three lines in general position. Their points of intersection have coordinates $(4 ; 3),(1; 1)$, and $(5; -1)$. Four solutions to the system are possible in the following tw... | Answer: $\left[\begin{array}{l}a=8, b \in N, b \neq 11, b \neq 39 \\ a=1, b \in N, b \neq 4, b \neq 11 \\ a=-2, b=1\end{array}\right.$ | =8,b\inN,b\neq11,b\neq39\\=1,b\inN,b\neq4,b\neq11\\=-2,b=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,518 |
1. For which natural values of $a$ does the equation $\gcd(4|\cos x|, a|\cos 3x|)=2$ have solutions? Find these solutions. | $$
\left[\begin{array}{l}
a=4 n-2, n \in N ; x=\frac{\pi k}{3}, k \in Z \\
a=4 n, n \in N ; x= \pm \frac{\pi}{3}+\pi k, k \in Z
\end{array}\right.
$$ | [\begin{pmatrix}=4n-2,n\inN;\frac{\pik}{3},k\inZ\\=4n,n\inN;\\frac{\pi}{3}+\pik,k\inZ\end{pmatrix} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,519 |
5. For what integers $a$ and $b$ does the system of equations
$$
\left\{\begin{array}{c}
|a x+2 y+b|=1 \\
(x-2 y+4)(5 x-y-7)(x+y+1)=0
\end{array}\right.
$$
have six solutions? | Answer: $\left\{\begin{array}{l}a \neq 1 ; a \neq-10 ; a \neq 2 ; \\ b \neq-2 a-5 ; b \neq-2 a-7 ; b \neq 2 a-1 ; \\ b \neq 2 a-3 ; b \neq 5-a ; b \neq 3-a\end{array}\right.$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,521 |
5. For which integers $a$ and $b$ does the system of equations
$$
\left\{\begin{array}{c}
|a x+b y+2|=1 \\
(3 x-5 y-2)(5 x+y-22)(x+3 y+4)=0
\end{array}\right.
$$
have five solutions? | Answer: $\left[\begin{array}{l}a=1-b, b \in Z, b \neq 1 \\ a=3-b, b \in Z, b \neq 2 \\ a=3 n+1, b=5 n+2, n \in Z, n \neq 0 \\ a=3 n, b=5 n+1, n \in Z, n \neq 0\end{array}\right.$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,524 |
1. For which natural values of $a$ does the equation $\gcd(4|\cos x|, a|\cos 2x|)=2$ have solutions? Find these solutions. | Answer: $\left[\begin{array}{l}a=4 n-2, n \in N ; x=\pi k, k \in Z \\ a=4 n, n \in N ; x= \pm \frac{\pi}{3}+\pi k, k \in Z\end{array}\right.$ | =4n-2,n\inN;\pik,k\inZ\\=4n,n\inN;\\frac{\pi}{3}+\pik,k\inZ | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 2,525 |
4. Students $10^{2}$ calculated the arithmetic means $x_{n}=\left(a_{1}+a_{2}+\ldots+a_{n}\right) / n$ of the terms of the numerical sequence $a_{k}=(4 k-3) / 5, k=1,2, \ldots, n, n=1,2, \ldots, 13$. The probability of making an error in calculating $x_{n}$ is proportional to $n$, and the event "to make an error in cal... | Answer: $\frac{46}{169}$ | \frac{46}{169} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,527 |
5. For which integers $a$ and $b$ does the system of equations
$$
\left\{\begin{array}{c}
|a x-2 y+b|=3 \\
(3 x-4 y-1)(4 x-y-10)(x+5 y+6)=0
\end{array}\right.
$$
have an infinite number of solutions? | Answer: $a=8, b=-17 ; a=8 ; b=-23$. | =8,b=-17;=8,b=-23 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,528 |
1. Answer: $1+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}=\frac{2648}{2145}$ | Definition of the progression: $p_{n}(2)=8 a_{n}-4\left(a_{n}+1\right)-2\left(6 a_{n}-1\right)+6=-8 a_{n}+4=-4-16 n \rightarrow a_{n}=1+2 n$
Factorization of the polynomial $p_{n}(x)$:
Checking the divisors of the free term:
$p_{n}(-2)=-8 a_{n}-4\left(a_{n}+1\right)+2\left(6 a_{n}-1\right)+6=0$, i.e., $x=-2$ is a ro... | \frac{2648}{2145} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,529 |
2. Answer: 1) $x_{n}=\frac{3\left(2^{n}-1\right)}{2^{n}}$ b 2) $\lim _{n \rightarrow \infty} x_{n}=3$
Option 0
For each natural number $n_{\text {write a formula for solving } x_{n}}$ of the equation $F_{n}(x)=\underbrace{f(f(f \ldots(f(x)) \ldots)}_{n}=0$, where the function $F_{n}(x)$ is the $n-$ fold composition o... | Answer: 1) $\left.x_{n}=-\frac{b\left(a^{n}-1\right)}{a^{n}(a-1)}, 2\right) \lim _{n \rightarrow \infty} x_{n}=-\frac{b}{a-1}$
Solution.
$$
\begin{aligned}
& F_{1}(x)=a x+b, F_{2}(x)=a(a x+b)+b=a^{2} x+b(a+1) \\
& F_{3}(x)=a\left(a^{2} x+b(a+1)\right)+b=a^{3} x+b\left(a^{2}+a+1\right), \ldots \\
& F_{n}(x)=a^{n} x+b\... | x_{n}=-\frac{b(^{n}-1)}{^{n}(-1)},\lim_{narrow\infty}x_{n}=-\frac{b}{-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,530 |
6. Let's introduce the following notations: $a$ - the side of the base, $b$ - the lateral edge, $N$ - the midpoint of side $AB$, $M$ - the midpoint of side $AD$, $K$ - the midpoint of edge $SC$ of the regular quadrilateral pyramid $SABCD$.
![](https://cdn.mathpix.com/cropped/2024_05_06_f36d8ea35c99389d52e3g-05.jpg?hei... | Answer: $S=\frac{5 \sqrt{2}}{8}$.
# | \frac{5\sqrt{2}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,533 |
6. Through the midpoints of sides $A B$ and $A D$ of the base of a regular quadrilateral pyramid $S A B C D$, a plane is drawn parallel to the median of the lateral face $S D C$, drawn from vertex $D$. Find the area of the section of the pyramid by this plane, if the side of the base of the pyramid is $\sqrt{2}$, and t... | Answer: $S=\frac{5}{2}$.
# | \frac{5}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,534 |
6. Through the midpoints of sides $A B$ and $A D$ of the base of a regular quadrilateral pyramid $S A B C D$, a plane is drawn parallel to the median of the lateral face $S D C$, drawn from vertex $D$. Find the area of the section of the pyramid by this plane, if the side of the base of the pyramid is 2, and the latera... | Answer: $S=5 \sqrt{2}$. | 5\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,535 |
6. Through the midpoints of sides $A B$ and $A D$ of the base of a regular quadrilateral pyramid $S A B C D$, a plane is drawn parallel to the median of the lateral face $S D C$, drawn from vertex $D$. Find the area of the section of the pyramid by this plane, if the side of the base of the pyramid is 2, and
the later... | Answer: $S=\frac{15 \sqrt{2}}{4}$.
Criteria for checking works, 11th grade
Preliminary round of the sectoral physics and mathematics olympiad for school students "Rosatom", mathematics, Moscow
# | \frac{15\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,536 |
5. Let point $M$ divide edge $A B$ in the ratio $A M: M B=\lambda$, point $N$ divide edge $D C$ in the ratio $D N: N C=\mu$, and point $P$ divide edge $D B$ in the ratio $D P: P B=\theta$. We need to find the ratio $A Q: Q C$.
 $\left\{\begin{array}{l}x_{1}=\cos \varphi \\ y_{1}=\sin \varphi\end{array}, \varphi \in\left[0 ; \frac{\pi}{2}\right]\right.$ when $\left.a=0 ; 2\right)\left\{\begin{array}{l}x_{1}=-\sin \frac{a}{2} \\ y_{1}=\cos \frac{a}{2}\end{array},\left\{\begin{array}{l}x_{2}=\cos \frac{a}{2} \\ y_{2}=-\sin \frac{a}{2}... | \cdot | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,541 |
5. In an arbitrary triangular pyramid $A B C D$, a section is made by a plane intersecting the edges $A B, D C$, and $D B$ at points $M, N, P$ respectively. Point $M$ divides edge $A B$ in the ratio $A M: M B=1: 2$. Point $N$ divides edge $D C$ in the ratio $D N: N C=3: 2$. Point $P$ divides edge $D B$ in the ratio $D ... | Answer: $A Q: Q C=3: 8$.
# | AQ:QC=3:8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,542 |
4. For which values of $a$ is the system $\left\{\begin{array}{c}x^{2}+y^{2}=1 \\ 3 \arcsin x-2 \arccos y=a\end{array}\right.$ consistent? | Answer: $a \in\left[-\frac{5 \pi}{2} ; \frac{\pi}{2}\right]$. | \in[-\frac{5\pi}{2};\frac{\pi}{2}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,544 |
5. In an arbitrary triangular pyramid $A B C D$, a section is made by a plane intersecting the edges $A B, D C$, and $D B$ at points $M, N, P$ respectively. Point $M$ divides edge $A B$ in the ratio $A M: M B=1: 3$. Point $N$ divides edge $D C$ in the ratio $D N: N C=4: 3$. Point $P$ divides edge $D B$ in the ratio $D ... | Answer: $A Q: Q C=4: 27$.
# | AQ:QC=4:27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,545 |
5. In an arbitrary triangular pyramid $A B C D$, a section is made by a plane intersecting the edges $A B, D C$, and $D B$ at points $M, N, P$ respectively. Point $M$ divides edge $A B$ in the ratio $A M: M B=2: 3$. Point $N$ divides edge $D C$ in the ratio $D N: N C=3: 5$. Point $P$ divides edge $D B$ in the ratio $D ... | Answer: $A Q: Q C=1: 10$.
Grading criteria, 11th grade
Preliminary round of the sectoral physics and mathematics school students' Olympiad "Rosatom", mathematics, trip 1
# | AQ:QC=1:10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,547 |
1. Let \( a \) be the length of \( AD \), \( x \) be the length of segment \( CD \), \( BD \) be perpendicular to line \( AD \), \( d \) be the length of the perpendicular, and \( \varphi \) be the angle of triangle \( CBD \). Let the speed in the field be \( v \). Then the speed on the road is \( 2v \). The travel tim... | Answer: $120^{\circ}$, if $\frac{d}{\sqrt{3}} \in[0 ; a] ; \pi-\arccos \frac{a}{\sqrt{a^{2}+d^{2}}}$, if $\frac{d}{\sqrt{3}}>a$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 2,548 |
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