problem stringlengths 0 45.3k | solution stringlengths 0 26.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
values | question_type stringclasses 5
values | problem_is_valid stringclasses 5
values | solution_is_valid stringclasses 5
values | source stringclasses 11
values | synthetic bool 2
classes |
|---|---|---|---|---|---|---|---|---|
5. Represent the number 2020 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: $2024=337^{3}+(335)^{3}+(-336)^{3}+(-336)^{3}+2^{3}$. | 2024=337^{3}+(335)^{3}+(-336)^{3}+(-336)^{3}+2^{3} | Number Theory | proof | Yes | Problem not solved | olympiads | false |
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 |
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 |
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. For what values of $a$ is the point with coordinates $(\cos a ; \sin a)$ symmetric to the point with coordinates $(\cos 2a ; -\sin 2a)$ with respect to the line with the equation $x-y=0$? | Answer: $a=-\frac{\pi}{2}+2 \pi n, n \in Z$. | -\frac{\pi}{2}+2\pin,n\inZ | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
4. Points $P, Q$ are located on the sides $A B$ and $A C$ of triangle $A B C$ such that $A P: P B=1: 4, A Q: Q C=3: 1$. 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 |
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 |
1. Petya considered moments of his life to be happy when his electronic watch showed the number of hours four times greater than the number of minutes or vice versa. Petya fell asleep and woke up at happy moments of his life, and he did not miss a single such moment. What is the maximum whole number of minutes Petya's ... | Answer: 299 minutes or 241 minutes. | 299 | Number Theory | math-word-problem | Yes | Incomplete | olympiads | false |
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 |
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. 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 |
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 |
1. Petya considered moments of his life to be happy when his electronic watch showed the number of hours six times greater than the minutes or vice versa. Petya fell asleep and woke up at happy moments of his life, and he did not miss a single such moment. What is the maximum whole number of minutes Petya's sleep could... | Answer: 423 minutes or 361 minutes. | 423 | Number Theory | math-word-problem | Yes | Incomplete | olympiads | false |
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 |
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 |
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
$$
\te... | Answer: $\left\{\begin{array}{l}a=15 k, k \in N ; \\ x= \pm \arccos \frac{3}{5}+\frac{\pi n}{2}, n \in Z .\end{array}\right.$ | {\begin{pmatrix}=15k,k\inN;\\\\arccos\frac{3}{5}+\frac{\pin}{2},n\inZ0\end{pmatrix}.} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Since $a^{2}-2 a+3=(a-1)^{2}+2$, then for $a \neq 1$ the value $a^{2}-2 a+3>2$ and the equation has no solutions, therefore the desired radius $R=\infty$. For $a=1$ the equation is equivalent to the system
$$
\left\{\begin{array}{l}
\sin (x+y-z)=1 \\
\cos (x+y-2 z)=1
\end{array}\right.
$$
Solving the system, we ge... | Answer: $\left\{\begin{array}{l}R=\infty, \quad a \neq 1 \\ R=\frac{\pi}{\sqrt{2}}, \quad a=1\end{array}\right.$. | R=\infty,\quad\neq1\\R=\frac{\pi}{\sqrt{2}},\quad=1 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
3. Since the parabola intersects the x-axis at $x=1$, the value of the function $y=a x^{2}+b x+c$ at this point is zero, i.e.,
$$
a+b+c=0
$$
The equation of the given line can be written as
$$
y=5-2 x
$$
The condition for the line to be tangent to the parabola at the point $(x ; y)$ is that the values of the functi... | Answer: $\left(-\frac{3}{5} ; \frac{16}{3}\right),\left(\frac{5}{3} ; \frac{4}{3}\right)$. | (-\frac{1}{3};\frac{16}{3}),(\frac{5}{3};\frac{4}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The first equation of the system defines a pair of parallel lines when $\mathrm{b} \neq 0$ and one line when $b=0$. The second equation defines three lines in general position. Their points of intersection have coordinates $(4 ; 3),(1; 1)$, and $(5; -1)$. Four solutions to the system are possible in the following tw... | Answer: $\left[\begin{array}{l}a=8, b \in N, b \neq 11, b \neq 39 \\ a=1, b \in N, b \neq 4, b \neq 11 \\ a=-2, b=1\end{array}\right.$ | =8,b\inN,b\neq11,b\neq39\\=1,b\inN,b\neq4,b\neq11\\=-2,b=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. For which natural values of $a$ does the equation $\gcd(4|\cos x|, a|\cos 3x|)=2$ have solutions? Find these solutions. | $$
\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.
$$ | [\begin{pmatrix}=4n-2,n\inN;\frac{\pik}{3},k\inZ\\=4n,n\inN;\\frac{\pi}{3}+\pik,k\inZ\end{pmatrix} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The coordinates $(x ; y ; z)$ of points $M$ in space are solutions to the equation $\cos (x+2 y-2 z)+\sin (y+z)+a^{2}+4 a+6=0$. The maximum radius of a sphere in space that does not contain such points inside. | Answer: $\frac{\pi \sqrt{22}}{6}$ | \frac{\pi\sqrt{22}}{6} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
5. For what integers $a$ and $b$ does the system of equations
$$
\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.
$$
have six solutions? | Answer: $\left\{\begin{array}{l}a \neq 1 ; a \neq-10 ; a \neq 2 ; \\ b \neq-2 a-5 ; b \neq-2 a-7 ; b \neq 2 a-1 ; \\ b \neq 2 a-3 ; b \neq 5-a ; b \neq 3-a\end{array}\right.$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The coordinates $(x ; y ; z)$ of points $M$ in space are solutions to the equation
$\sin (x-2 y+2 z)+\sin (2 x+2 y+z)=|a+3|+2$.
Find the
maximum radius of a sphere in space that does not contain such points inside. | Answer: $\frac{\pi \sqrt{2}}{3}$ | \frac{\pi\sqrt{2}}{3} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. The line with the equation $4x + y - 7 = 0$ is tangent to the parabola $y = ax^2 + bx + c$ at a point with integer coordinates. Find the coordinates of the point of tangency, given that the parabola intersects the $x$-axis at the point with abscissa $x = 2$, and the numbers $a, b, c$ are integers. | Answer: $(3 ;-5),(1 ; 3)$ | (3,-5),(1,3) | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
5. For which integers $a$ and $b$ does the system of equations
$$
\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.
$$
have five solutions? | Answer: $\left[\begin{array}{l}a=1-b, b \in Z, b \neq 1 \\ a=3-b, b \in Z, b \neq 2 \\ a=3 n+1, b=5 n+2, n \in Z, n \neq 0 \\ a=3 n, b=5 n+1, n \in Z, n \neq 0\end{array}\right.$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. For which natural values of $a$ does the equation $\gcd(4|\cos x|, a|\cos 2x|)=2$ have solutions? Find these solutions. | Answer: $\left[\begin{array}{l}a=4 n-2, n \in N ; x=\pi k, k \in Z \\ a=4 n, n \in N ; x= \pm \frac{\pi}{3}+\pi k, k \in Z\end{array}\right.$ | =4n-2,n\inN;\pik,k\inZ\\=4n,n\inN;\\frac{\pi}{3}+\pik,k\inZ | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The coordinates $(x ; y ; z)$ of points $M$ in space are solutions to the equation
$$
\sin (x+y+2 z)+\cos (x-3 y+z)+|a-2|+2=0
$$
Find the maximum radius of a sphere in space that does not contain such points inside it. | Answer: $\frac{\pi \sqrt{1122}}{66}$ | \frac{\pi\sqrt{1122}}{66} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
4. Students $10^{2}$ calculated the arithmetic means $x_{n}=\left(a_{1}+a_{2}+\ldots+a_{n}\right) / n$ of the terms of the numerical sequence $a_{k}=(4 k-3) / 5, k=1,2, \ldots, n, n=1,2, \ldots, 13$. The probability of making an error in calculating $x_{n}$ is proportional to $n$, and the event "to make an error in cal... | Answer: $\frac{46}{169}$ | \frac{46}{169} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. For which integers $a$ and $b$ does the system of equations
$$
\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.
$$
have an infinite number of solutions? | Answer: $a=8, b=-17 ; a=8 ; b=-23$. | =8,b=-17;=8,b=-23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Answer: $1+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}=\frac{2648}{2145}$ | Definition of the progression: $p_{n}(2)=8 a_{n}-4\left(a_{n}+1\right)-2\left(6 a_{n}-1\right)+6=-8 a_{n}+4=-4-16 n \rightarrow a_{n}=1+2 n$
Factorization of the polynomial $p_{n}(x)$:
Checking the divisors of the free term:
$p_{n}(-2)=-8 a_{n}-4\left(a_{n}+1\right)+2\left(6 a_{n}-1\right)+6=0$, i.e., $x=-2$ is a ro... | \frac{2648}{2145} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Answer: 1) $x_{n}=\frac{3\left(2^{n}-1\right)}{2^{n}}$ b 2) $\lim _{n \rightarrow \infty} x_{n}=3$
Option 0
For each natural number $n_{\text {write a formula for solving } x_{n}}$ of the equation $F_{n}(x)=\underbrace{f(f(f \ldots(f(x)) \ldots)}_{n}=0$, where the function $F_{n}(x)$ is the $n-$ fold composition o... | Answer: 1) $\left.x_{n}=-\frac{b\left(a^{n}-1\right)}{a^{n}(a-1)}, 2\right) \lim _{n \rightarrow \infty} x_{n}=-\frac{b}{a-1}$
Solution.
$$
\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 |
4. Answer: 1) $n=-4+24 t \quad 2) n=-8+24 t, t \in Z$ | Solution.
Notation: $t=\sin \frac{5 \pi n}{12}$
Investigation of the function $f(t)=6 t+\left(3 t-4 t^{3}\right)=9 t-4 t^{3}, t \in[-1 ; 1]$
Critical points:
$f^{\prime}(t)=9-12 t^{2}=0 \rightarrow\left[\begin{array}{c}t_{1}=\frac{\sqrt{3}}{2} \\ t_{2}=-\frac{\sqrt{3}}{2}\end{array}\right]$. At the point $t=t_{1}$ ... | -4+24t\quad | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
5. Answer: $d_{\min }=\frac{r-l}{2}=1$ | Solution
The trajectories of the midpoints of segments $M N$ and $P Q$ when their endpoints traverse the boundary of the square are shown in Fig. 1 and Fig. 2:
Fig 1
![](https://cdn.mathpix... | d_{\}=\frac{r-}{2}=1 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
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 |
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 |
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. The number of addends in the expression is even. Let's break them into pairs of adjacent addends and sum them within each pair:
$$
\underset{2 k}{1122 \ldots . .211-1122 \ldots . . .211}=101 \cdot 10^{2 k+1}, k=1,2, \ldots, 1010
$$
The sum of the pairs (geometric progression):
$$
\begin{aligned}
& 101 \cdot 10^{3... | Answer: $-10 \underbrace{20202 \ldots 02}_{504 \text{ digits}} 01000$. | -10\underbrace{20202\ldots02}_{504digits}01000 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
4. Parameterize the solutions of the first equation of the system $\left\{\begin{array}{l}x=\cos \varphi \\ y=\sin \varphi\end{array}, \forall \varphi \in[0 ; 2 \pi)\right.$ and substitute into the second equation $\arcsin (\cos \varphi)+\arccos (\sin \varphi)=a$. The graph below shows the dependence of $a$ on $\varphi... | Answer: For $a=0\left\{\begin{array}{l}x=\cos \varphi \\ y=\sin \varphi\end{array}, \forall \varphi \in\left[\frac{\pi}{2} ; \pi\right]\right.$. For $a=\pi,\left\{\begin{array}{l}x=\cos \varphi \\ y=\sin \varphi\end{array}, \forall \varphi \in\left[\frac{3 \pi}{2} ; 2 \pi\right]\right.$. | notfound | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
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 $A Q: Q C$.
. | Answer: $-\underbrace{1212 \ldots 1200}_{1010 \text{ digits }}$ | -\underbrace{1212\ldots1200}_{1010 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
4. Solve the system $\left\{\begin{array}{c}x^{2}+y^{2}=1 \\ \arccos x-\arcsin y=a\end{array}\right.$ for any $a$. | Answer: 1) $\left\{\begin{array}{l}x_{1}=\cos \varphi \\ y_{1}=\sin \varphi\end{array}, \varphi \in\left[0 ; \frac{\pi}{2}\right]\right.$ 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 |
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. Calculate $143-1443+14443-\ldots-144 \ldots 43$ (the signs alternate).
2020 | Answer: $\underbrace{1313 . .1300}_{1010 \text { triples of the number }}$.
| 131300.1300 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
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 |
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. Calculate $154-1554+15554-\ldots-155 \ldots 54$ (the signs alternate). 2020 | Answer: $\underbrace{1414 \ldots .1400}_{1010 \text { chepverokiedinii }}$ | \underbrace{1414\ldots0.1400}_{1010 | Algebra | math-word-problem | Incomplete | Incomplete | olympiads | false |
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 |
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. The number $a=2^{m} \cdot 3^{n}, m \geq 1, n \geq 1$ has a total of $(m+1)(n+1)$ divisors, and from the condition it follows that $m+1=2^{k_{1}} \cdot 3^{k_{2}}, n+1=2^{s_{1}} \cdot 3^{s_{2}}, k_{1}+s_{1} \leq m, k_{2}+s_{2} \leq n$. We have the following inequalities:
$$
\begin{gathered}
a \leq 1000 \rightarrow 2^... | Answer: 1) 11 numbers; 2) $a_{\max }=972$.
| | | $n=1$ | $n=2$ | $n=3$ | $n=5$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| | $n+1$ | 2 | 3 | 4 | 6 |
| $m=1$ | 2 | 6 | 18 | 54 | 486 |
| $m=2$ | 3 | 12 | 36 | 108 | 972 |
| $m=3$ | 4 | 24 | 72 | 216 | 1944 |
| $m=5$ | 6 | 96 | 288 | 864 | 7776 |
| $m=7$ | 8 ... | 972 | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
5. In triangle $ABC$ with area $s$, medians $AM, BN, CQ$ are drawn. A point $P$ is taken on line $AC$ such that $AP: PC = \lambda$.
Case 1. Point $P$ is located on side $AC$.
$$
\begin{aligned}
& \frac{AP}{PC} = \lambda \rightarrow AP = \frac{\lambda}{\lambda+1} AC \rightarrow PN = \frac{|\lambda-1|}{2(\lambda+1)} AC... | Answer: $S_{1}=3, S_{2}=6$.
# | notfound | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
1. Village A is located on a straight section of asphalt road, and Village B is in the field, 2 km away from the road. Petya decided that he would walk part of the way along the road and then turn off it and go straight to B. His speed on the road is 5 km/h, and in the field - 3 km/h. Find the length of the path Petya ... | Answer: 2.5 km, if $a \geq 1.5$ km ; $\sqrt{4+a^{2}}$, if $a<1.5$ km. | 2.5 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
2. A positive integer $a \leq 2000$ has only two prime divisors: 2 and 5, and the number of all its divisors, including 1 and $a$, is itself a divisor of $a$. How many such numbers $a$ exist? Find the smallest among them. | Answer: 1) 5 numbers; 2) $a_{\min }=40$. | 5;a_{\}=40 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $ABC$ with an area of 10, medians $AM, BN, CQ$ are drawn. On line $AC$, a point $P$ is taken such that $AP: PC = 3: 2$. Find the area of the quadrilateral with vertices at points $P, M, N, Q$. | Answer: $S_{1}=3, S_{2}=15$.
# | notfound | Geometry | math-word-problem | Yes | Problem not solved | olympiads | false |
2. A positive integer $a \leq 10^{5}$ has only two prime divisors: 3 and 5, and the number of all its divisors, including 1 and $a$, is itself a divisor of $a$. How many such numbers $a$ exist? Find the largest among them. | Answer: 1) 4 numbers; 2) $a_{\max }=50625$. | 4;a_{\max}=50625 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $ABC$ with an area of 20, medians $AM, BN, CQ$ are drawn. On line $AC$, a point $P$ is taken such that $AP: PC = 2: 3$. Find the area of the quadrilateral with vertices at points $P, M, N, Q$. | Answer: $S_{1}=6, S_{2}=30$.
# | 6 | Geometry | math-word-problem | Yes | Problem not solved | olympiads | false |
2. A positive integer $a \leq 50000$ has only two prime divisors: 2 and 7, and the number of all its divisors, including 1 and $a$, is itself a divisor of $a$. How many such numbers $a$ exist? Find the smallest among them. | Answer: 1) 5 numbers; 2) $a_{\text {min }}=56$. | 5;a_{}=56 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $ABC$ with an area of 14, medians $AM, BN, CQ$ are drawn. On line $AC$, a point $P$ is taken such that $AP: PC = 4: 3$. Find the area of the quadrilateral with vertices at points $P, M, N, Q$. | Answer: $S_{1}=4, S_{2}=28$.
Criteria for checking works, 11th grade
Preliminary round of the sectoral physics and mathematics olympiad for schoolchildren "Rosatom", mathematics, field 2
# | 4 | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
1. Let $[x]$ and $\{x\}$ be the integer and fractional parts of the number $x$. The integer part of the number $x$ is the greatest integer not exceeding $x$, and $\{x\}=x-[x]$. Find $x$ if $2 x+3[x]-5\{x\}=4$. | Solution.
$x=[x]+\{x\} \rightarrow 2([x]+\{x\})+3[x]-5\{x\}=4 \rightarrow 5[x]-3\{x\}=4 \rightarrow\{x\}=\frac{5[x]-4}{3} \in[0 ; 1)$
$[x] \in\left[\frac{4}{5} ; \frac{7}{5}\right) \rightarrow[x]=1 \rightarrow\{x\}=\frac{1}{3} \rightarrow x=[x]+\{x\}=1+\frac{1}{3}=\frac{4}{3}$
Answer: $x=\frac{4}{3}$. | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Represent the number 80 as the sum of two prime numbers. In how many ways can this be done? Let's remind you that one is not considered a prime number.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution. We organize the enumeration of all prime numbers from 2 to 73 in the form of a table.
| 1 | 7 | 13 | 19 | 25 | 31 | 37 | 43 | 49 | 55 | 61 | 67 | 73 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2 | 8 | 14 | 20 | 26 | 32 | 38 | 44 | 50 | 56 | 62 | 68 | 74 |
... | 4 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
4. On a sheet of paper, 12 consecutive integers are written. After one of them is crossed out, the sum of the remaining numbers equals 325. Which number was crossed out? | Solution. Let $n, n+1, \ldots, n+k-1, n+k, n+k+1, \ldots, n+11$ be 12 consecutive integers, and the number $n+k, k=0,1,2, \ldots, 11$ is crossed out. The sum of the numbers after crossing out is
$$
\frac{2 n+11}{2} \cdot 12-(n+k)=325 \rightarrow 11 n+66-k=325 \rightarrow k=11 n-259
$$
Considering the condition $k \in... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. You have a cardboard circle with a radius of 4, a ruler, and a compass. On a sheet of paper, construct a rectangle equal in area to the circle. | Solution. Construction.
Let's draw on a plane using a compass and a ruler two mutually perpendicular lines
Fig.
and a circle of radius \( R \), tangent to the horizontal line, with its cen... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 5. Construction.
## 3 variant | Answer: $x=\frac{11}{4} ; \frac{7}{2} ; \frac{17}{4} ; 5$. | \frac{11}{4};\frac{7}{2};\frac{17}{4};5 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Task 1. Answer: 22 (23, if the start is considered) .
# | # Solution.
Case 1. The first and second cars are located on the same diameter. This happens at times $t_{1,2}^{k}$ (min), when the distance between the cars along the track is a multiple of half its length.
$$
t_{1,2}^{k}=\frac{0.375}{3} \cdot 60 \cdot k=7.5 \cdot k, k=1,2, \ldots, 8
$$
Case 2. The first and third ... | 22 | Other | math-word-problem | Incomplete | Yes | olympiads | false |
Task 2. Answer: 250.
# | # Solution.
$$
A=(5 n+1) 3^{n}-7=(5 n+1)(4-1)^{n}-7=4 B+(-1)^{n}(5 n+1)-7
$$
for some integer $B$. Then the divisibility of $A$ by 4 means the divisibility by 4 of the number
$$
(-1)^{n}(5 n+1)-7
$$
Case $1 . n=2 k, k=1,2, \ldots 500$.
$$
\begin{aligned}
& (-1)^{n}(5 n+1)-7=10 k-6=4 m \rightarrow 5 k-2 m=3 \righta... | 250 | Other | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 4. Answer: triples formed by permuting the numbers ( $11 ; 2 ; 13$ ) and triples obtained by permuting the numbers $(11 ; 3 ; 7)$ (a total of 12 different triples). | Solution. One of the numbers $(p ; q ; r)$ is divisible by 11, so due to its simplicity, for example, $p$ equals 11. Then
$$
q r=q+r+11 \rightarrow r(q-1)=q+11 \rightarrow q \neq 1 \rightarrow r=\frac{q+11}{q-1}=1+\frac{12}{q-1} .
$$
Case 1. $q-1=1 \rightarrow q=2, r=13 \rightarrow(p ; q ; r)=(11 ; 2 ; 13)$.
Case 2.... | (11;2;13) | Combinatorics | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 5. Answer: $O M: O C=13: 6$.
## Variant 0
Points $M$ and $N$ divide the sides $A D$ and $C D$ of parallelogram $A B C D$ in the ratios $A M: M D=\alpha: \beta$ and $C N: N D=\lambda: \mu$, respectively. Lines $C M$ and $B N$ intersect at point $O$. Find the ratio of the lengths of segments $O M$ and $O C$. Fi... | Answer: 1) $O M: O C=\frac{\beta}{\alpha+\beta}+\frac{\mu}{\lambda} ; \quad$ 2) $B O: O N=\frac{(\lambda+\mu)(\alpha+\beta)}{\lambda \beta}$.
## Solution.
Notations: $A M=\alpha x, M B=\bet... | Geometry | math-word-problem | Yes | Yes | olympiads | false | |
Problem 3. Answer: $\frac{1}{64}$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided translation is already in the requested format and directly represents the translation of the given text. However, the not... | Solution. The coordinates of each pair of intersection points of the line and the hyperbola satisfy the system of equations:
$$
\left\{\begin{array}{c}
y=\frac{k}{x} \\
y=2 k x+b
\end{array} \rightarrow \frac{k}{x}=2 k x+b \rightarrow 2 k x^{2}+b x-k=0\right.
$$
The product of the abscissas of the two intersection po... | \frac{1}{64} | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Answer: $B O: O N=4$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the note is to be considered part of the translation requ... | Solution. In the case $2 \alpha=1, \beta=3, \lambda=1, \mu=2$. By formula (*) we get $B O: O N=4$
## Variant 3 | BO:ON=4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Answer: $O M: O C=33: 7$. | Solution. In option $3 \alpha=2, \beta=5, \lambda=1, \mu=4$. From formula (**) we get $O M: O C=33: 7$.
## Option 4 | OM:OC=33:7 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 5. Answer: $B O: O N=35: 8$. | Solution. In the case $4 \alpha=1, \beta=4, \lambda=2, \mu=5$. By formula (*) we get $B O: O N=35: 8$. | BO:ON=35:8 | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
# 2.16. Qualifying round of the "Rosatom" Olympiad, 7th grade | # 2.16. Qualifying round of the "Rosatom" Olympiad, 7th grade
## Answers and solutions
## Problem 1
$222222232323=22 \cdot\left(10^{4}+10^{2}+1\right) \cdot 10^{6}+23 \cdot\left(10^{4}+10^{2}+1\right)$
$242424=24 \cdot\left(10^{4}+10^{2}+1\right)$
$\frac{222222232323}{242424}=\frac{22 \cdot 10^{6}+23}{24}=\frac{22... | \frac{733341}{8} | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
1. There are 38 students in the class. One of them has 16 red pencils, 24 blue pencils, 32 green pencils, and 40 black pencils. This student decided to give each of his classmates a set of three pencils of different colors. Will he be able to carry out his plan? What is the maximum number of such sets he can assemble? | Task 1.
1) An incorrect answer is given without justifications - 0 points;
2) The solution is carried out, but the answer is incorrect - 0.5 points;
3) The answer is correct, but poorly justified (without an example) - 1.5 points. | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In a notebook, several positive numbers were written. Each of them is equal to one-sixth of the sum of the others. How many numbers are written in the notebook? | Task 2.
1) There is a solution, but the answer is incorrect - 0.5 points;
2) The answer is correct, but it is not proven that all numbers are the same - 1 point;
3) The answer is correct, but the justification is insufficient - 1.5 points; | Algebra | math-word-problem | Yes | Yes | olympiads | false | |
3. For a win in a chess tournament, a participant receives one point, for a draw - half a point, and for a loss - 0 points. Petya played 18 games in the tournament and scored $10 \frac{1}{2}$ points. By how many games did he win more than he lost? | Task 3.
1) There is a solution, but the answer is incorrect - 0.5 points;
2) The answer is correct, but there is no justification - 0.5 points;
3) The answer is correct, but the justification is poor - 1 point;
4) The answer is obtained by trial and error - 1.5 points. | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false | |
4. How many different ordinary irreducible fractions of the form $\frac{253}{p}$ exist in the interval $\left(\frac{1}{254} ; \frac{1}{253}\right) ?$ | Task 4.
1) There is a solution, but the answer is incorrect - 0.5 points;
2) The answer is incorrect because reducible fractions were not considered - 0.5 points;
3) The answer is insufficiently justified - 1.5 points.
Task 5.
1) On the graph, the streets intersect - 0 points;
2) The graph is drawn with a gap - 0.5 ... | notfound | Number Theory | math-word-problem | Yes | Problem not solved | olympiads | false |
2. the sum decreases when $c$ increases with fixed $a$ and $b$ and permissible (integer) $d$.
Case $1 . a=6 \rightarrow b \in[7 ; 12]$. For all $b \in[7 ; 11]$ there exist permissible $c$ and $d$. When $b=12 \rightarrow c<\frac{5 \cdot 18}{8}=11.25$ and permissible $c$ do not exist. $\sum_{\text {min }}=42$ is achieve... | Solution.
The circle has its center at the point $\left(1 ; \frac{5}{2}\right)$ and radius $r=\frac{\sqrt{5}}{2}$. It touches the lines $y=2(x-1)$ and $y=-2(x-1)$ at the points $(2 ; 4)$ and ... | \in(\frac{-11-\sqrt{5}}{4};-\frac{5}{2}]\cup(\frac{\sqrt{5}-17}{8};\frac{-\sqrt{5}-7}{8})\cup[-\frac{1}{2};\frac{\sqrt{5}-1}{4}) | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
5. A football is sewn from 256 pieces of leather: white ones in the shape of hexagons and black ones in the shape of pentagons. Black pentagons only border white hexagons, and any white hexagon borders three black pentagons and three white hexagons. Find the number of white hexagons on the football. | Answer: 160.
Criteria for checking works, 7th grade
Preliminary round of the sectoral physics and mathematics olympiad for schoolchildren "Rosatom", mathematics
# | 160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Edges $A B$ and $C D$ of the triangular pyramid $A B C D$ are mutually perpendicular and equal in length to 4. The lengths of edges $B C$ and $B D$ are 3 and 2, and the lengths of edges $A C$ and $A D$ are $2 \sqrt{3}$ and $\sqrt{17}$, respectively. Find the volume of the pyramid.
## Answers and solutions
Problem ... | Solution
The number $a \in E_{f}$, if the equation $\frac{7 x^{2}+22 x+16}{x^{2}+2 x+2}=a$ has a solution:
$x^{2}+2 x+2 \neq 0 \rightarrow 7 x^{2}+22 x+16=a x^{2}+2 a x+2 a \rightarrow(a-7) x^{2}+2(a-11) x+2 a-16=0$.
For $a \neq 7$, the quadratic equation has a solution if its discriminant is non-negative:
$D / 4=(... | \frac{\sqrt{71}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Lines $l_{1}, l_{2}$, and $l_{3}$ on a plane are parallel, the distances between lines $l_{1}, l_{2}$ and $l_{2}, l_{3}$ are 1 and 2 respectively (line $l_{2}$ is located between lines $l_{1}$ and $l_{3}$). The vertices of an equilateral triangle $A B C$ lie on lines $l_{1}, l_{2}$, and $l_{3}$. Find the length of t... | Solution
When $y=0 \rightarrow x=0$. For $y \neq 0$, divide the left side of the equation $x^{2}+x y^{2}-10 y^{2}=0$ by $y^{2}$:
After substituting $t=\frac{x}{y}$, we get $t^{2}+x-10=0 \rightarrow x=10-t^{2} \rightarrow y=\frac{x}{t}=-t+\frac{10}{t}$. The number $t^{2}$ must be an integer. We will prove that $t$ is ... | notfound | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
8. $p=-10 \rightarrow\left\{\begin{array}{c}x=-90 \\ y=9\end{array}\right.$
Problem 2 Answer: $\frac{7}{4}$ | Transformation: $4 \sin \frac{x-y}{2} \cdot \cos \frac{x+y}{2}=2 \sin \frac{x-y}{2} \cdot \cos y \rightarrow \sin \frac{x-y}{2}\left(2 \cos \frac{x+y}{2}-\cos y\right)=0$.
1 case. $\sin \frac{x-y}{2}=0 \rightarrow x=y+\pi k, k \in Z$.
Then $F=\sin ^{2} \frac{x+y}{2}+\left.\cos y\right|_{x=y+2 \pi k}=\sin ^{2} y+\cos ... | \frac{7}{4} | Algebra | math-word-problem | More than one problem | Incomplete | olympiads | false |
6. At point A of a sphere with center at point $o$ and radius 4, there is a light source.
On the plane passing through point $o$ and perpendicular to the line $A O$, inside the sphere, there is a target, as shown in the figure. Point $o$ on the target coincides with the center of the sphere, and the radii $\rho$ and $... | Solution
Transformation: $x(x(x+8)+12)=\frac{a}{x+8}$.
Notation: $t=x(x+8) \in[-16 ;+\infty), t \neq 0$. Then $a=t^{2}+12 t$. The minimum value of $a$ is achieved at $t=-6 \in[-16 ;+\infty)$ and equals -36.
Problem 2 Answer: 1) $f_{\max }=11 \quad 2) x=\operatorname{arctg} 3+\pi k, k \in Z$
Solution
When $\cos x=0... | \frac{5696\pi}{425} | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
6. Points $M$ and $N$ are the midpoints of the edges $AD$ and $BC$ of the base of a regular quadrilateral pyramid $SABCD$ with a side length of the base 2 and the apothem of the lateral face $\sqrt{3}$. On the apothem of the lateral face $ASD$, there is a point $P$ such that $MP: PS = 1: 2$. On the apothem of the later... | Solution
Let's look for a function among polynomials of the second degree: $f(x)=a x^{2}+b x+c$ with some coefficients $a, b, c$.
$$
f(x+1)=a(x+1)^{2}+b(x+1)+c=f(x)+2 a x+a+b \rightarrow
$$
$$
\rightarrow f(x+1)-f(x)=2 a x+a+b \equiv x+2
$$
Then $\left\{\begin{array}{c}2 a=1 \\ a+b=2\end{array}\right.$ and $a=\frac... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. $n=2 k, k \in Z$
$m=\frac{1}{2}\left(4 k^{2}+8 a_{1} k-2 k+2-4 a_{1}\right)=2 k^{2}-k+1+2 a_{1}(k-1)$ will be an integer for any $k$, if $2 a_{1}-$ is an integer.
Combining the two cases, we get $a_{1}=\frac{t}{2}, \quad t \in Z$.
Let's find $t \in Z$, for which $m \geq n$ :
$n^{2}+\left(4 a_{1}-1\right) n-4 a_{... | Solution
We will use the property of GCD: $\forall a, b, c \in N \rightarrow \text{GCD}(a, b) = \text{GCD}(a, b + ac) \cdot$
Case 1. $x > y$.
GCD $\left(x+y, x^{2}+2 y^{2}\right) = \text{GCD} \left(x+y, x^{2}+2 y^{2}+2(x+y)(x-y) = \text{GCD} \left(x+y, 3 x^{2}\right)\right)$
If $p \geq 4$ is a common divisor of the... | 3 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
2. The left part of the equation can be represented as a difference of squares:
$$
\begin{aligned}
& (\sin x \cdot \operatorname{ctg} 2 x+1)^{2}-\left(\sin x-\operatorname{ctg}^{2} 2 x\right)^{2}=0 \Leftrightarrow(\sin x \cdot \operatorname{ctg} 2 x+1)^{2}=\left(\sin x-\operatorname{ctg}^{2} 2 x\right)^{2} \Leftrighta... | Answer: $x=-\frac{\pi}{6}+\frac{2 \pi}{3} k, k \in Z, k \neq 3 t+1, t \in Z ; \quad x=\frac{\pi}{10}+\frac{2 \pi}{5} m, m \in Z, m \neq 5 t+1, t \in Z$. | -\frac{\pi}{6}+\frac{2\pi}{3}k,k\inZ,k\neq3+1,\inZ;\quad\frac{\pi}{10}+\frac{2\pi}{5},\inZ,\neq5+1,\inZ | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
3. We will factorize the differences of cubes in the numerator and the sums of cubes in the denominator:
$$
A=\frac{(2-1)\left(2^{2}+2+1\right)}{(2+1)\left(2^{2}-2+1\right)} \cdot \frac{(3-1)\left(3^{2}+3+1\right)}{(3+1)\left(3^{2}-3+1\right)} \cdot \cdots \cdot \frac{(99-1)\left(99^{2}+99+1\right)}{(99+1)\left(99^{2}... | Answer: $\quad \frac{2\left(100^{2}+100+1\right)}{3 \cdot 100 \cdot 101}=\frac{3367}{5050}$. | \frac{3367}{5050} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let's write $x$ in the form: $x=[x]+\alpha$, where $\alpha$ is the fractional part of the number $x, \quad \alpha \in[0 ; 1)$. Then the equation $x^{3}-2[x]=5$ will take the form: $x^{3}-2(x-\alpha)=5$ or $2 \alpha=5+2 x-x^{3}$.
Since $2 \alpha \in[0 ; 2)$, we will find the possible integer values of $x$ that satis... | Answer: $x_{1}=\sqrt[3]{7}, x_{2}=\sqrt[3]{9}$. | x_{1}=\sqrt[3]{7},x_{2}=\sqrt[3]{9} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
8. We found that in the isosceles triangle $P Q R$, the angle at the vertex is $60^{\circ}$, which means it is equilateral. Then $S_{P Q R}=\frac{\sqrt{3}}{4} P R^{2}=\sqrt{3}$. | Answer: $\quad S_{P Q R}=\sqrt{3}$.
# | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\cos ^{2} x \cdot \operatorname{tg}^{4} 2 x-\cos ^{2} x-\operatorname{tg}^{4} 2 x+4 \cos x \cdot \operatorname{tg}^{2} 2 x+1=0$. | Answer: $\quad x=\frac{\pi}{3}+\frac{2 \pi}{3} k, k \in Z ; \quad x=\frac{\pi}{5}+\frac{2 \pi}{5} n, n \in Z ; \quad x=2 \pi m, m \in Z$. | \frac{\pi}{3}+\frac{2\pi}{3}k,k\inZ;\quad\frac{\pi}{5}+\frac{2\pi}{5}n,n\inZ;\quad2\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the value of the product $\frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \frac{4^{3}-1}{4^{3}+1} \cdot \ldots \cdot \frac{200^{3}-1}{200^{3}+1}$. | Answer: $\frac{2 \cdot\left(200^{2}+200+1\right)}{3 \cdot 200 \cdot 201}=\frac{40201}{60300}$. | \frac{40201}{60300} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $x^{3}-3[x]=4$. Here $[x]$ is the integer part of the number $x$ - the greatest integer not exceeding $x$. | Answer: $\quad x_{1}=-\sqrt[3]{2}, \quad x_{2}=\sqrt[3]{7}, \quad x_{3}=\sqrt[3]{10}$. | x_{1}=-\sqrt[3]{2},\quadx_{2}=\sqrt[3]{7},\quadx_{3}=\sqrt[3]{10} | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
5. On the sides $A B$ and $B C$ outside the triangle $A B C$, two equilateral triangles $A B M$ and $B C N$ are constructed. Points $P, Q$, and $R$ are the midpoints of segments $A B, M N$, and $B C$ respectively. Find the area of triangle $P Q R$, if the length of side $A C$ of triangle $A B C$ is 8. | Answer: $\quad S_{P Q R}=4 \sqrt{3}$. | 4\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\cos ^{2} x \cdot \operatorname{ctg}^{4} 2 x-\cos ^{2} x-\operatorname{ctg}^{4} 2 x+4 \cos x \cdot \operatorname{ctg}^{2} 2 x+1=0$. | Answer: $\quad x=\frac{2 \pi}{3} k, k \in Z, k \neq 3 t, t \in Z ; \quad x=\frac{2 \pi}{5} n, n \in Z, n \neq 5 t, t \in Z$. | \frac{2\pi}{3}k,k\inZ,k\neq3,\inZ;\quad\frac{2\pi}{5}n,n\inZ,n\neq5,\inZ | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
3. Calculate the value of the product $\frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \frac{4^{3}-1}{4^{3}+1} \cdot \ldots \cdot \frac{300^{3}-1}{300^{3}+1}$. | Answer: $\quad \frac{2 \cdot\left(300^{2}+300+1\right)}{3 \cdot 300 \cdot 301}=\frac{90301}{135450}$. | \frac{90301}{135450} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $x^{3}-6[x]=7$. Here $[x]$ is the integer part of the number $x$ - the greatest integer not exceeding $x$. | Answer: $\quad x_{1}=-\sqrt[3]{11}, \quad x_{2}=-\sqrt[3]{5}, \quad x_{3}=\sqrt[3]{19}$. | x_{1}=-\sqrt[3]{11},\quadx_{2}=-\sqrt[3]{5},\quadx_{3}=\sqrt[3]{19} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the sides $A B$ and $B C$ outside the triangle $A B C$, two equilateral triangles $A B M$ and $B C N$ are constructed. Points $P, Q$, and $R$ are the midpoints of segments $A B, M N$, and $B C$ respectively. Find the area of triangle $P Q R$, if the length of side $A C$ of triangle $A B C$ is 2. | Answer: $\quad S_{P Q R}=\frac{\sqrt{3}}{4}$.
# | \frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\sin ^{2} x \cdot \operatorname{tg}^{4} 2 x-\sin ^{2} x-\operatorname{tg}^{4} 2 x+4 \sin x \cdot \operatorname{tg}^{2} 2 x+1=0$. | Answer: $\quad x=\frac{\pi}{6}+\frac{2 \pi}{3} k, k \in Z ; \quad x=-\frac{\pi}{10}+\frac{2 \pi}{5} n, n \in Z ; \quad x=\frac{\pi}{2}+2 \pi m, m \in Z$. | \frac{\pi}{6}+\frac{2\pi}{3}k,k\inZ;\quad-\frac{\pi}{10}+\frac{2\pi}{5}n,n\inZ;\quad\frac{\pi}{2}+2\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Calculate the value of the product $\frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \frac{4^{3}-1}{4^{3}+1} \cdot \ldots \cdot \frac{400^{3}-1}{400^{3}+1}$. | Answer: $\frac{2 \cdot\left(400^{2}+400+1\right)}{3 \cdot 400 \cdot 401}=\frac{53467}{80200}$. | \frac{53467}{80200} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $x^{3}-9[x]=8$. Here $[x]$ is the integer part of the number $x$ - the greatest integer not exceeding $x$. | Answer: $\quad x_{1}=-\sqrt[3]{28}, \quad x_{2}=-\sqrt[3]{19}, x_{3}=-1, \quad x_{4}=\sqrt[3]{26}, \quad x_{5}=\sqrt[3]{35}$. | x_{1}=-\sqrt[3]{28},\quadx_{2}=-\sqrt[3]{19},\quadx_{3}=-1,\quadx_{4}=\sqrt[3]{26},\quadx_{5}=\sqrt[3]{35} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
5. On the sides $A B$ and $B C$ outside the triangle $A B C$, two equilateral triangles $A B M$ and $B C N$ are constructed. Points $P, Q$, and $R$ are the midpoints of segments $A B, M N$, and $B C$ respectively. Find the area of triangle $P Q R$, if the length of side $A C$ of triangle $A B C$ is 1. | Answer: $\quad S_{P Q R}=\frac{\sqrt{3}}{16}$.
Grading Criteria, 10th Grade
## Preliminary Round of the ROSATOM Industry Physics and Mathematics Olympiad for Schoolchildren, Mathematics
# | \frac{\sqrt{3}}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.