problem stringlengths 0 45.3k | solution stringlengths 0 26.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
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6. Planes $P$ and $Q$, parallel to the base of a regular quadrilateral pyramid $S A B C D$, intersect the edge $S A$ of the pyramid at points $M$ and $N$. The lengths of the segments $S M, S N$, and $S A$ are three consecutive terms of a geometric progression with a common ratio $q=3$. Find the dihedral angle at the ba... | Solution
Transformation: $a>0, a \neq 1, x>0$
$2 \log _{a}^{2} x+3 \log _{a} x-3=\frac{2}{\log _{a} x} \rightarrow 2 \log _{a}^{3} x+3 \log _{a}^{2} x-3 \log _{a} x-2=0$
Substitution $t=\log _{a} x$ leads to the cubic equation $2 t^{3}+3 t^{2}-3 t-2=0$, which has three roots
$t_{1}=1, t_{2}=-2, t_{3}=-\frac{1}{2}$.... | notfound | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
6. The plane $P$ intersects the lateral edges $S A, S B, S C$ of the triangular pyramid $S A B C$ at points $M, N, K$ respectively and forms an angle of $45^{\circ}$ with the lateral face $S B C$. Find the volume of the pyramid $S A B C$, if the product of its edges $S A \cdot S B \cdot S C=5 \sqrt{15}$, and the pyrami... | Solution
Let's find the general term of the sequence:
$a_{2}=2 \cdot a+3 \rightarrow a_{3}=2(2 \cdot a+3)+3=2^{2} \cdot a+2 \cdot 3+3 \rightarrow a_{4}=2\left(2^{2} \cdot a+2 \cdot 3+3\right)+3=2^{3} \cdot a+3\left(2^{2}+1\right)$
$a_{n}=2^{n-1} \cdot a+3\left(2^{n-2}+2^{n-1}+\ldots+1\right)=2^{n-1} \cdot a+3\left(2... | notfound | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
2. $s_{1}=4, s_{2}=2 \rightarrow a_{2}=3^{4} \cdot 5^{2}=2025$
Problem 4 Answer: $P(A)=\frac{3}{4}$ | Solution
Notations: $x, y, z$ - measures (in degrees) of the arcs of the circle on which the angles at the vertices $A, B, C$ of the triangle are based.
By the condition $x+y+z=360, x>0, y>... | P(A)=\frac{3}{4} | Number Theory | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 6. $V_{\text {SABC }}=\frac{p \sqrt{3} \cos ^{2} \alpha \sin \alpha}{\left(1+3 \cos ^{2} \alpha\right)^{3 / 2}}=\frac{p \sqrt{3} \operatorname{tg} \alpha}{\left(\operatorname{tg}^{2} \alpha+4\right)^{3 / 2}}$. In option 3. $\alpha=30^{\circ}, p=13 \sqrt{39}, V=9$
Option 4
Problem 1 Answer: $\boldsymbol{a}=446... | Solution
Transformation:
$\sqrt{3}(\sin x+\cos x-\sqrt{2})+2 \sin 2 x(\sin x+\cos x-\sqrt{2})=0 \rightarrow$
$(\sin x+\cos x-\sqrt{2})(2 \sin 2 x+\sqrt{3})=0$
Problem 3 Answer: 295245
Pr... | Algebra | MCQ | More than one problem | Yes | olympiads | false | |
6. In a regular quadrilateral pyramid, opposite lateral faces are perpendicular. The height of the pyramid is $h$. Find the radius of the sphere that touches the edges of the base and the lateral edges of the pyramid or their extensions.
Variant 1
Problem 1 Answer: $x=\frac{\pi}{6}+2 \pi k, k \in Z$ | notfound | Geometry | math-word-problem | Incomplete | Problem not solved | olympiads | false | |
1. The class was divided into two teams, "Know-alls" and "Don't-know-alls," and they started playing a "Questions and Answers" game. The rules of the game are simple: each player on a team must be able to ask a question on a chosen topic to any player on the other team and answer a question addressed to them by an oppo... | 1. Task
- Correctly wrote down the system -- 0.5 points.
- Started solving correctly, but made a mistake (answer is incorrect) -- 1.0
- Correct approach to the solution, but incorrect answer or minor errors -- 1.5
- Fully correct solution and correct answer -- 2.0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | |
2. Solve the equation $f(x)+g(x)=0$, for $f(x)=\sin x+\sin 3 x+\sin 5 x+\ldots+\sin 2025 x$ and $g(x)=\cos x+\cos 3 x+\cos 5 x+\ldots+\cos 2025 x$. | 2. Task
- Correctly guessed to multiply by $\sin \mathrm{x}$, or started adding into sums of two terms with a constant sum of arguments, but did not progress further, incorrect solution
$-0.5$
- Found one series of solutions (not all)
$--1.0$
- Found both series, but did not discard $\sin x=0$, or minor errors
$-... | notfound | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
3. How many different triples of natural numbers $a, b, c$ exist such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{4}$? (triples differing in the order of their elements are considered different) | 3. Task
- Saw several particular cases (less than 4, not all) -- 0.5
- Found the correct restrictions on a (the smallest number), did not progress further, there are errors
- Found all correct sets (a, b, c), error in calculating permutations, there are errors
$--1.5$
- Fully correct solution and correct answer
$--... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. How many triangles exist with a perimeter of 50, where the lengths of the sides are integers? | 4. Task
- Correctly wrote down the conditions for the ratios of the sides of the triangle
- Obtained the correct system of inequalities for the sides, but did not solve it
- Correct approach to the solution, but there are minor errors
$--1.5$
- Fully correct solution and correct answer
$--2.0$ | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The base $AD$ of parallelogram $ABCD$ is divided by points $M_{1}, M_{2}, \ldots, M_{6}$ into seven equal parts. The lines $B M_{1}, B M_{2}, \ldots B M_{6}$ intersect the diagonal $AC$ at points $N_{1}, N_{2}, \ldots, N_{6}$ respectively. Find the length of the fourth segment from vertex $A$ in the division of the ... | 5. Task
- Drew the correct diagram, saw the similarity of triangles, or began applying Menelaus' theorem, but with little progress
- Correctly wrote some proportions for the sides, but there is no complete solution --1.0
- Made minor computational errors in an otherwise correct solution
$--1.5$
- Fully correct solut... | Geometry | math-word-problem | Yes | Yes | olympiads | false | |
1. The arithmetic progression $a_{n}$ is such that its difference is not equal to zero, and $a_{10}, a_{13}$, and $a_{19}$ are consecutive terms of some geometric progression. Find the ratio $a_{12}: a_{18}$. | Solution. Let $a_{n}=a_{1}+d(n-1)$, where $a_{1}$ is the first term of the progression, and $d$ is its common difference.
Then $a_{10}=a_{1}+9 d, a_{13}=a_{1}+12 d, a_{19}=a_{1}+18 d$. By the property of a geometric progression, we have $a_{10} \cdot a_{19}=a_{13}^{2}:\left(a_{1}+9 d\right)\left(a_{1}+18 d\right)=\lef... | 5:11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. How many solutions does the equation $\left[2 x-x^{2}\right]+2\{\cos 2 \pi x\}=0$ have? Indicate the smallest and the largest of them. Here $[a]$ is the integer part of the number $a$ - the greatest integer not exceeding $a$, $\{a\}=a-[a]$ is the fractional part of the number $a$. | Solution. Rewrite the equation as $\left[2 x-x^{2}\right]=-2\{\cos 2 \pi x\}$. Since
$$
\{\cos 2 \pi x\} \in[0 ; 1) \rightarrow\left[2 x-x^{2}\right] \in(-2 ; 0] \rightarrow\left[2 x-x^{2}\right]=-1,\left[2 x-x^{2}\right]=0
$$
There are two possible cases:
Case 1. $\left\{\begin{array}{l}{\left[2 x-x^{2}\right]=-1} ... | x_{\}=-\frac{1}{3},x_{\max}=\frac{7}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. For which $a$ does the system $\left\{\begin{array}{l}x \cos a+y \sin a=5 \cos a+2 \sin a \\ -3 \leq x+2 y \leq 7,-9 \leq 3 x-4 y \leq 1\end{array}\right.$ have a unique solution? | Solution. The inequalities in the system limit a parallelogram:
$$
y=\frac{-x-3}{2}, y=\frac{-x+7}{2}, y=\frac{3 x+9}{4}, y=\frac{3 x-1}{4} \text {. }
$$
Let's find its vertices:
$$
\begin{aligned}
& A:\left\{\begin{array}{l}
x+2 y+3=0 \\
3 x-4 y-1=0
\end{array} \rightarrow A(-1 ;-1) \quad B:\left\{\begin{array}{c}
... | a_{1}=\operatorname{arctg}4+\pik,a_{2}=-\operatorname{arctg}2+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. For what least integer $n$ are all solutions of the equation
$$
x^{3}-(5 n-9) x^{2}+\left(6 n^{2}-31 n-106\right) x-6(n-8)(n+2)=0 \text { greater than }-1 \text { ? }
$$ | Solution. Rewrite the equation as
$$
x^{3}-(5 n-9) x^{2}+\left(6 n^{2}-31 n-106\right) x-6 n^{2}+36 n+96=0 .
$$
Notice that $x_{1}=1>-1$ is a solution for all $n$ (the sum of the coefficients of the polynomial on the left side of the equation is zero for all $n$). Divide the equation by $x-1$:
$$
x^{2}-5(n-2) x+6\le... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The midline of a trapezoid is 4. A line parallel to the bases of the trapezoid and dividing its area in half intersects the lateral sides at points $M$ and $N$. Find the smallest possible length of the segment $M N$. | # Solution.
Notations: $A D=a, B C=b, C K=H, N E=h$ - heights of triangles $N C P$ and $D N Q$, $M N=x$
$$
S_{\text {AMND }}=S_{\text {MBCN }} \rightarrow \frac{x+b}{2} H=\frac{a+x}{2} h \r... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A convex quadrilateral $A B C D$ is inscribed in a circle. Its diagonal $B D$ is the bisector of the angle at vertex $B$, forms an acute angle of $72^{0}$ with the other diagonal and an angle of $53^{0}$ with side $A D$. Find the angles of the quadrilateral.
Problem 1 Answer: 3136. | Solution.
A
$\mathrm{D}$
On each side of the square, there are 10 points (including the vertices). The vertices of an arbitrary convex quadrilateral can be 8 points on each side of the squ... | notfound | Geometry | math-word-problem | Incomplete | Incomplete | olympiads | false |
3. Pete, Vanya, Sergey, and Ivan Ivanovich are the founders of an enterprise with some initial capital. If Pete doubles his share in the initial capital, it will increase by $30 \%$. If Vanya does the same, the capital will increase by $25 \%$. If Sergey triples his share, the capital growth will be $50 \%$. By what fa... | Answer: more than 6 times. | than6times | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
5. A convex quadrilateral $A B C D$ is inscribed in a circle. Its diagonal $B D$ is the bisector of the angle at vertex $B$, forms an acute angle of $80^{\circ}$ with the other diagonal and an angle of $55^{\circ}$ with side $A D$. Find the angles of the quadrilateral. | Answer: $80^{0}, 90^{0}, 100^{0}, 90^{0}$ or $100^{0}, 50^{0}, 80^{0}, 130^{0}$.
## Variant 3 | 80^{0},90^{0},100^{0},90^{0} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Pete, Vanya, Sergey, and Ivan Ivanovich are the founders of an enterprise with some initial capital. If Pete doubles his share in the initial capital, it will increase by $25 \%$. If Vanya does the same, the capital will increase by $35 \%$. If Sergey triples his share, the capital growth will be $40 \%$. By what fa... | Answer: more than 4 times | than4times | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
5. A convex quadrilateral $A B C D$ is inscribed in a circle. Its diagonal $B D$ is the bisector of the angle at vertex $B$, forms an acute angle of $75^{\circ}$ with the other diagonal and an angle of $70^{\circ}$ with side $A D$. Find the angles of the quadrilateral. | Answer: $75^{\circ}, 70^{0}, 105^{0}, 110^{0}$ or $105^{\circ}, 10^{0}, 75^{\circ}, 170^{0}$.
## Variant 4 | 75,70,105,110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Pete, Vanya, Sergey, and Ivan Ivanovich are the founders of an enterprise with some initial capital. If Pete doubles his share in the initial capital, it will increase by $30 \%$. If Vanya does the same, the capital will increase by $40 \%$. If Sergey triples his share, the capital growth will be $20 \%$. By what fa... | Answer: more than 12 times. | than12times | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
5. A convex quadrilateral $A B C D$ is inscribed in a circle. Its diagonal $B D$ is the bisector of the angle at vertex $B$, forms an acute angle of $65^{\circ}$ with the other diagonal and an angle of $55^{\circ}$ with side $A D$. Find the angles of the quadrilateral. | Answer: $65^{\circ}, 120^{\circ}, 115^{0}, 60^{0}$ or $115^{\circ}, 20^{\circ}, 65^{0}, 160^{0}$.
## Grading Criteria for the Final Round of the Rosatom Olympiad 04.03.2023 8th Grade
In all problems, the correct answer without justification 0 | 65,120,115,60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Answer: $k=16$ collisions.
A wire in the shape of a circle with radius $R$ has $m$ beads strung on it, equally spaced from each other. At a certain moment, $(m-1)$ beads are made to move with the same speed $v(1/$ sec) in the counterclockwise direction, while the remaining bead is made to move in the opposite ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | |
4. There is a stick 15 cm long and a rope, the length of which is more than 1 m. You need to measure the length of the rope with an accuracy of 5 cm using the stick. You are allowed to: 1) lay the stick along the rope an integer number of times; 2) lay pieces of the rope, shorter than 15 cm, along the stick. You are no... | # Solution
## First Algorithm
Connect the beginning and the end of the rope to form a closed loop.
Step 1. Lay the stick along the rope from point A, for example, counterclockwise until the remaining piece of the rope is less than 15 cm (point $A_{1}$).
 35 numbers, 2) 10003
Problem 3 Answer: the exponent is 2003.
Problem 4 Answer:
1) $n=4 t+1, t=0,1,2, \ldots$
2) 250
Problem 5 Answer: yes, the side... | Solution: Let the side length of the triangles in the partition be $\frac{1}{n}$ of the side length of the given triangle. Then, in the first strip, there is 1 triangle, in the second strip, there are 2 more, i.e., 3 triangles, in the third strip, there are 2 more, i.e., 5, and so on. In the $k$-th strip, there are $2 ... | yes,thesideofthesoughttriangleis\frac{1}{45}ofthesideofthegiventriangle | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Answer: the distance of point C from the mouth of the tributary is not less than 4 km. | Notation: $a, b$ are the distances of points $A, B$ and $C$ from the mouth of the tributary, $v$ is the own speed of the kayak, $w$ is the speed of the current, $T_{k}$ is the time of travel on the route, $k$ is the route number. There are six possible route variants:
Route 1. $A \rightarrow B \rightarrow C$
$$
T_{1}... | notfound | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 4. Answer: $\alpha=(-1)^{n} \frac{\pi}{6}+\pi n, n \in Z ; \beta=\frac{\pi}{3}+2 \pi k, k \in Z$. | Solution. We find the coordinates of the points of intersection of the circle and the line:
$$
\begin{aligned}
& \left\{\begin{array} { c }
{ x ^ { 2 } + y ^ { 2 } = 1 } \\
{ 2 x + 2 y - 1 - \sqrt { 3 } = 0 }
\end{array} \rightarrow \left\{\begin{array}{c}
x=\cos \varphi, y=\sin \varphi \\
\sin \varphi+\cos \varphi=\... | \alpha=(-1)^{n}\frac{\pi}{6}+\pin,n\inZ;\beta=\frac{\pi}{3}+2\pik,k\inZ | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 5. Answer: $B D=3$.
Option 0
In triangle $A B C$, the length of side $A C$ is $2 b$, side $A B-c$, and the angle at vertex $A$ is $60^{\circ}$. Point $M$, located on median $B E$, divides it in the ratio $B M: M E=p: q$. Line $A M$ intersects side $B C$ at point $D$. Find the length of segment $B D$. | Answer: $B D=\frac{p}{p+2 q} \cdot \sqrt{4 b^{2}+c^{2}-2 b c}$
## Solution.
By the Law of Cosines, we find the length of side $B C: B C^{2}=4 b^{2}+c^{2}-2 b c$. Menelaus' Theorem:
, D\left(\frac{\sqrt{2}}{2} ; \frac{\sqrt{3}}{2}\right)$.
Notation: $A(\cos \beta ; \sin 2 \alpha), B(\sin 3 \alpha ; \sin 2 \beta)$
Case 1. $A=C, B=D$
$$
\left\{\begin{array} { c }
{ \operatorname { cos } \beta = \frac { 1 }... | \alpha=\frac{\pi}{4}+2\pin,n\inZ,\beta=\frac{\pi}{3}+2\pik,k\inZ | Algebra | math-word-problem | Not a problem | Yes | olympiads | false |
Problem 4. Answer: $\left\{\begin{array}{l}\alpha= \pm \frac{\pi}{6}+2 \pi n, n \in Z \\ \beta= \pm \frac{\pi}{4}+2 \pi k, k \in Z\end{array},\left\{\begin{array}{l}\alpha= \pm \frac{\pi}{4}+2 \pi n, n \in Z \\ \beta= \pm \frac{\pi}{6}+2 \pi k, k \in Z\end{array}\right.\right.$ | Solution. Points of intersection of the circle and the line: $C\left(\frac{1}{2} ; \frac{\sqrt{2}}{2}\right), D\left(0 ; \frac{\sqrt{3}}{2}\right)$.
Notation: $A(\cos 2 \alpha ; \cos \beta), B(\cos 2 \beta ; \cos \alpha)$.
Case 1. $A=C, B=D$
$$
\left\{\begin{array} { l }
{ \operatorname { cos } 2 \alpha = \frac { 1... | {\begin{pmatrix}\alpha=\\frac{\pi}{6}+2\pin,n\inZ\\\beta=\\frac{\pi}{4}+2\pik,k\inZ\end{pmatrix},{\begin{pmatrix}\alpha=\\frac{\pi}{4}+2\pin,n\inZ\\\beta=\\frac} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 4. Answer: $\left\{\begin{array}{c}\alpha= \pm \frac{2 \pi}{3}+2 \pi n, n \in Z \\ \beta=\frac{\pi}{4}+\frac{\pi k}{2}, k \in Z\end{array}\right.$ | Solution. Points of intersection of the circle and the line: $C\left(-\frac{1}{2} ; 0\right), D\left(0 ;-\frac{1}{2}\right)$.
Notation: $A(\cos 2 \alpha ; \cos 2 \beta), B(\cos 2 \beta ; \cos \alpha)$
Case 1. $A=C, B=D$
$$
\left\{\begin{array} { l }
{ \operatorname { c o s } 2 \alpha = - \frac { 1 } { 2 } } \\
{ \o... | {\begin{pmatrix}\alpha=\\frac{2\pi}{3}+2\pin,n\inZ\\\beta=\frac{\pi}{4}+\frac{\pik}{2},k\inZ\end{pmatrix}.} | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Problem 5. Answer: $B D=\frac{14}{5}$. | Solution.
$b=4, c=5, p=4, q=3 \rightarrow \sqrt{4 b^{2}+c^{2}-2 b c}=\sqrt{64+25-40}=7 \rightarrow BD=\frac{4}{4+6} \cdot 7=\frac{14}{5}$. | \frac{14}{5} | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
6. A wolf is surrounded by dogs located at points $M, N, P$ and $Q$ on the sides of a square $A B C D$, $M \in[A ; B], N \in[B ; C], P \in[C ; D], Q \in[D ; A]$ such that $A M: M B=B N: N C=C P: P D=D Q: Q A=1: 3$. The wolf, located inside the square at the intersection of lines $M P$ and $N Q$, can run in a straight l... | Solution
If $\varphi \in[0 ; 2 \pi]$, then the numbers $x=\frac{5}{2} \cos \varphi$ and $y=\frac{5}{3} \sin \varphi$ satisfy the equation. Then $x-2 y=\frac{5}{2} \cos \varphi-\frac{10}{3} \sin \varphi=\frac{5}{6}(3 \cos \varphi-4 \sin \varphi)=\frac{25}{6}\left(\frac{3}{5} \cos \varphi-\frac{4}{5} \sin \varphi\right)... | \frac{25}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Answer: $\left\{\begin{array}{l}x=\frac{5 \pi}{2}(5-4 t) \\ y=12 \pi(t-1)\end{array}, t \in Z\right.$ | Solution.
Since $|2 \sin (x+y)+3| \leq 5,|\cos (2 x-y)-1| \leq 2$, equality is possible only when $\left\{\begin{array}{c}\sin (x+y)=1 \\ \cos (2 x-y)=-1\end{array} \rightarrow\left\{\begin{array}{l}x+y=\frac{\pi}{2}+2 \pi m, \\ 2 x-y=\pi+2 \pi n\end{array} \rightarrow\left\{\begin{array}{l}x=\frac{\pi}{2}+\frac{2 \pi... | notfound | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
6. The hare found itself cornered by two dogs in a circle with a radius of 40 and is located at point $A$, which is 20 units away from the center of the circle $O$. The dogs are at the ends of the diameter of the circle passing through points $A$ and $O$. They can pursue the hare only along the boundary of the circle a... | Solution. Substitution $u=2^{x}, \quad v=3^{y} \rightarrow 27 u^{2}-30 u v+8 v^{2}=0$. Substitution
$t=\frac{u}{v} \rightarrow 27 t^{2}-30 t+8=0 \rightarrow\left[\begin{array}{l}t=2 / 3 \\ t=4 / 9\end{array}\right.$.
Let $\frac{2^{x}}{3^{y}}=\frac{2}{3} \rightarrow 2^{x-1}=3^{y-1}$. If $x>1, y>1$, then there are no i... | \frac{\sqrt{5}}{\pi} | Geometry | math-word-problem | Incomplete | Yes | olympiads | false |
Task 1. Answer: 1) $y=16$
2) $f_{\max }=\frac{1}{12}$ | Solution
Let's switch to logarithms with base $x$ (in the case $1 x=2$): $\frac{\log _{x} y-1}{\left(\log _{x} y+2\right)^{2}}, x>0, x \neq 1, y>0$. Substituting the variable $u=\log _{x} y \rightarrow f(u)=\frac{u-1}{(u+2)^{2}}$. The graph of the function $f(u)$ is shown below:
 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
Task 3. Answer:
1) $x=7 n^{2}+6 n+1, n \in Z$
2) $x=7 n^{2}+8 n+2$ | Solution
$7 x+2=y^{2} \rightarrow 7 x=y^{2}-2$, i.e., the number $y^{2}-2$ must be divisible by 7.
Case 1. $y=7 n, n \in Z$. Then $7 x=49 n^{2}-2$. Such integer $x$ does not exist, since the right side of the equation is not divisible by 7. The same result is obtained for $y=7 n+1, y=7 n+2, \quad y=7 n+5, \quad y=7 n... | \frac{\sqrt{117+24\sqrt{14}}}{5} | Number Theory | math-word-problem | Incomplete | Incomplete | olympiads | false |
1. Find all $x$ that satisfy the inequality $n^{2} x^{2}-\left(2 n^{2}+n\right) x+n^{2}+n-6 \leq 0$ for any natural $n$. | Solution. The roots of the quadratic trinomial $n^{2} x^{2}-\left(2 n^{2}+n\right) x+n^{2}+n-6$ are
$$
x_{1}=1-\frac{2}{n}, x_{2}=1+\frac{3}{n}
$$
Factorize the left side of the inequality
$$
n^{2}\left(x-\left(1-\frac{2}{n}\right)\right)\left(x-\left(1+\frac{3}{n}\right)\right) \leq 0
$$
Solving the inequality usi... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $\left(\cos \frac{2 x}{5}-\cos \frac{2 \pi}{15}\right)^{2}+\left(\sin \frac{2 x}{3}-\sin \frac{4 \pi}{9}\right)^{2}=0$. | Solution. The equation is equivalent to the system of equations $\left\{\begin{array}{l}\cos \frac{2 x}{5}-\cos \frac{2 \pi}{15}=0 \\ \sin \frac{2 x}{3}-\sin \frac{4 \pi}{9}=0\end{array}\right.$.
Solve the equation $\cos \frac{2 x}{5}=\cos \frac{2 \pi}{15}:\left[\begin{array}{c}\frac{2 x}{5}=\frac{2 \pi}{15}+2 \pi k, ... | \frac{29\pi}{3}+15\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find $x$ and $y$ for which $\left\{\begin{array}{l}x-2 y+[x]+3\{x\}-\{y\}+3[y]=2.5 \\ 2 x+y-[x]+2\{x\}+3\{y\}-4[y]=12\end{array}\right.$, where $[x],[y]$ and $\{x\},\{y\}$ are the integer and fractional parts of the numbers $x$ and $y$. The integer part of a number $a$ is the greatest integer not exceeding $a$, and ... | Solution. Given that $x=[x]+\{x\}$ and $y=[y]+\{y\}$, we can rewrite the system as:
$$
\left\{\begin{array}{l}
2[x]+4\{x\}+[y]-3\{y\}=2.5 \\
{[x]+4\{x\}-3[y]+4\{y\}=12}
\end{array}\right.
$$
Introduce the notations: $\{x\}=u \in[0 ; 1),\{y\}=v \in[0 ; 1)$.
The system becomes: $\left\{\begin{array}{l}2[x]+[y]=2.5-4 u... | (\frac{53}{28};-\frac{23}{14});(\frac{5}{2};-\frac{3}{2});(\frac{87}{28};-\frac{19}{14}) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find the probability that a number randomly taken from the interval $[0 ; 5]$ is a solution to the equation $\sin (x+|x-\pi|)+2 \sin ^{2}(x-|x|)=0$.
# | # Solution.
Case 1. $x \in[0 ; \pi]$. The equation takes the form: $\sin (x+\pi-x)+2 \sin ^{2}(x-x)=0$, i.e., any $x \in[0 ; \pi]$ is a solution to the equation, therefore, $P(A)=\frac{\pi}{5}$.
Case $2, x \in(\pi ; 5]$. The equation takes the form: $\sin (x+x+\pi)+2 \sin ^{2}(x-x)=0$ or $\sin 2 x=0$. Then $x=\frac{3... | \frac{\pi}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. For which $a$ does the system of equations $\left\{\begin{array}{c}x \sin a-y \cos a=2 \sin a-\cos a \\ x-3 y+13=0\end{array}\right.$ have a solution $(x ; y)$ in the square $5 \leq x \leq 9,3 \leq y \leq 7 ?$ | Solution. The line $L$, defined by the equation $x-3 y+13=0$, intersects the sides of the square at points $A(5 ; 6)$ and $B(8 ; 7)$.
The first equation of the system can be written as $(x-... | \in[\frac{\pi}{4}+\pik,\operatorname{arctg}\frac{5}{3}+\pik],k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On the edges of a trihedral angle with vertex at point $S$, there are points $M, N$, and $K$ such that $S M^{2}+S N^{2}+S K^{2} \leq 12$. Find the area of triangle $S M N$, given that the angle $M S N$ is $30^{\circ}$, and the volume of the pyramid $S M N K$ is maximally possible. | Solution. Let's introduce the notations: $S M=m, S N=n, S K=k$.
The volume of the pyramid $S M N K$ is
$$
V=\frac{1}{3} S_{S M N} \cdot h=\frac{1}{6} m n \sin \alpha \cdot h=\frac{1}{6} m n... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Let $\{x\}$ and $[x]$ denote the fractional and integer parts of the number $x$. The integer part of the number $x$ is the greatest integer not exceeding $x$, and $\{x\}=x-[x]$. Find $x$ for which $4 x^{2}-5[x]+8\{x\}=19$. | Solution.
$$
\begin{aligned}
& x=[x]+\{x\} \rightarrow 4([x]+\{x\})^{2}-5[x]+8\{x\}=19 \rightarrow \\
& 4\{x\}^{2}+8([x]+1)\{x\}+4[x]^{2}-5[x]-19=0
\end{aligned}
$$
If $x$ is the desired value, then the quadratic trinomial $f(t)=4 t^{2}+8([x]+1) t+4[x]^{2}-5[x]-19$ has a root $t=\{x\}$ on the interval $[0 ; 1)$.
Cas... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the integers $n$ for which the expression $\frac{1}{12}\left(8 \sin \frac{\pi n}{10}-\sin \frac{3 \pi n}{10}+4 \cos \frac{\pi n}{5}+1\right)$ takes integer values. | Solution. Let's introduce the notation $t=\sin \frac{\pi n}{10} \in[-1 ; 1]$ and rewrite the original expression:
$$
\begin{aligned}
& \frac{1}{12}\left(8 \sin \frac{\pi n}{10}-\sin \frac{3 \pi n}{10}+4 \cos \frac{\pi n}{5}+1\right)=\frac{1}{12}\left(8 t-\left(3 t-4 t^{3}\right)+4\left(1-2 t^{2}\right)+1\right)= \\
& ... | 20k-5,k\inZ | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. On a plane, there are 8 lines, 3 of which are parallel, and any two of the remaining five intersect. We consider all triangles with sides lying on these lines. What is the greatest and the least number of such triangles that can be found? | Solution. Let's introduce the notation: $P$ - the set of parallel lines, $p_{k}, k=1,2, \ldots, m$ - any line from $P$; $Q$ - the set of non-parallel lines, $q_{j}, j=1,2, \ldots, n$ - any line from $Q$. The maximum possible number of triangles is associated with such an arrangement of lines from $Q$ where no three of ... | N_{\max}=40,N_{\}=20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A random variable $\xi$ is uniformly distributed on the interval $[0 ; 6]$. Find the probability that the inequality $x^{2}+(2 \xi+1) x+3-\xi \leq 0$ holds for all $x \in[-2 ;-1]$. | Solution. The inequality $f(x)=x^{2}+(2 \xi+1) x+3-\xi \leq 0$ holds on the interval $[-2 ;-1]$, if
$$
\left\{\begin{array} { l }
{ f ( - 1 ) \leq 0 } \\
{ f ( - 2 ) \leq 0 }
\end{array} \rightarrow \left\{\begin{array} { c }
{ 1 - ( 2 \xi + 1 ) + 3 - \xi \leq 0 } \\
{ 1 - 2 ( 2 \xi + 1 ) + 3 - \xi \leq 0 }
\end{arr... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. For what values of $a$ does the system $\left\{\begin{array}{c}(x-7 \cos a)^{2}+(y-7 \sin a)^{2}=1 \\ |x|+|y|=8\end{array}\right.$ have a unique solution? | Solution. The set of points on the plane, with coordinates $(x ; y)$ that satisfy the first equation of the system, lie on a circle $K_{a}$ of radius 1 centered at the point $O(7 \cos a ; 7 \sin a)$. As $a$ changes, the center moves along a circle of radius 7 centered at the origin.
\pi}{4},k\inZ | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. In a triangular pyramid $S A B C$, the angle $A S B$ at vertex $S$ is $30^{\circ}$, and the lateral edge $S C$ is inclined to the plane of the face $A S B$ at an angle of $45^{\circ}$. The sum of the lengths of the lateral edges of the pyramid is 9. Find the greatest possible value of the volume of the pyramid under... | # Solution.
Let's introduce the notations: $S A=a, S B=b, S C=c, C K=h, \square A S B=\alpha, \square C S K=\beta$,
$C K$ is the perpendicular to the plane $A S B$. Then
$$
V_{S A B C}=\fr... | \frac{9\sqrt{2}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Quadrilateral $ABCD$ is inscribed in a circle, and its diagonals intersect at point $P$. Points $K, L$, and $M$ are the midpoints of sides $AB, BC$, and $CD$ respectively. The radius of the circle circumscribed around triangle $KLP$ is 1. Find the radius of the circle circumscribed around triangle $LMP$.
Problem 1 ... | Solution. Let $y_{k}$ be the number of passengers in the car with number $k, k=1,2,3, \ldots, 10$. According to the problem, $\sum_{k=1}^{10} y_{k}=270$. Additionally, it is stated that $y_{2} \geq y_{1}+2, y_{3} \geq y_{1}+4, \ldots, y_{9} \geq y_{1}+16, \quad y_{10} \geq y_{1}+18$. Adding these inequalities,
$$
y_{2... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. There are 12 carriages in the train, and they contain 384 passengers. In the second carriage, there are more than two passengers more than in the first, in the third carriage, there are more than two passengers more than in the second, and so on until the last carriage. The number of passengers in the last carriage ... | Answer: There are no solutions.
Solution. Let $y_{k}$ be the number of passengers in the carriage with number $k, k=1,2,3, \ldots, 12$. According to the problem, $\sum_{k=1}^{12} y_{k}=384$. Additionally, it is stated that $y_{2} \geq y_{1}+3, y_{3} \geq y_{1}+6, \ldots, y_{11} \geq y_{1}+30, y_{12} \geq y_{1}+33$. Ad... | nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the largest solution of the equation on the interval $(0 ; 2 \pi)$
$$
(\sin x + \cos x + \sin 3x)^{3} = \sin^{3} x + \cos^{3} x + \sin^{3} 3x
$$ | Answer: $x=\frac{15 \pi}{8}$. | \frac{15\pi}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the first 1000 digits after the decimal point in the decimal representation of the number $(8+\sqrt{65})^{2022}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Answer: all 1000 digits after the decimal point are zeros. | 0 | Number Theory | math-word-problem | Yes | Incomplete | olympiads | false |
4. Solve the equation $\left|3 x-\sqrt{1-9 x^{2}}\right|=\sqrt{2}\left(18 x^{2}-1\right)$. | Answer: $x=\frac{\sqrt{2+\sqrt{3}}}{6} ; x=\frac{1}{3 \sqrt{2}}$. | \frac{\sqrt{2+\sqrt{3}}}{6};\frac{1}{3\sqrt{2}} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
2. Find the largest solution of the equation on the interval $(0 ; 2 \pi)$
$$
(\cos 2 x+\sin 3 x+\cos 4 x)^{3}=\cos ^{3} 2 x+\sin ^{3} 3 x+\cos ^{3} 4 x
$$ | Answer: $x=\frac{25 \pi}{14}$. | \frac{25\pi}{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the first 2000 digits after the decimal point in the decimal representation of the number $(9+4 \sqrt{5})^{2021}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Answer: all 2000 digits after the decimal point are nines. | 999\ldots999 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $\left|4 x-\sqrt{1-16 x^{2}}\right|=\left(32 x^{2}-1\right) \sqrt{1-16 x^{2}}$. | Answer: $x=\frac{1}{4 \sqrt{2}}$. | \frac{1}{4\sqrt{2}} | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
2. Find the largest solution of the equation on the interval $(0 ; 2 \pi)$
$$
(\cos 3 x+\cos 4 x+\cos 5 x)^{3}=\cos ^{3} 3 x+\cos ^{3} 4 x+\cos ^{3} 5 x
$$ | Answer: $x=\frac{17 \pi}{9}$. | \frac{17\pi}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the first 200 digits after the decimal point in the decimal representation of the number $(7+5 \sqrt{2})^{2020}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Answer: all 200 digits after the decimal point are zeros. | 0 | Combinatorics | math-word-problem | Yes | Incomplete | olympiads | false |
4. Solve the equation $\left|5 x-\sqrt{1-25 x^{2}}\right|=5 \sqrt{2} x\left(100 x^{2}-3\right)$. | Answer: $x=\frac{\sqrt{2+\sqrt{2}}}{10} ; x=-\frac{\sqrt{2-\sqrt{2}}}{10}$. | \frac{\sqrt{2+\sqrt{2}}}{10};-\frac{\sqrt{2-\sqrt{2}}}{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
0.35 \cdot 160+0.1 x=0.2 \cdot 160+0.2 x, 0.15 \cdot 160=0.1 x, x=240 \text {. }
$$
In total, it results in $160+240=400$ g of solution. | Answer: 400.
## Problem 9
Points $B_{1}$ and $C_{1}$ are the feet of the altitudes of triangle $ABC$, drawn from vertices $B$ and $C$ respectively. It is known that $AB=7, \quad AC=6, \sin \angle BAC=\frac{2 \sqrt{110}}{21}$. Find the length of the segment $B_{1}C_{1}$.
## Solution:
Let the angle $\angle BAC$ be de... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
0.6 \cdot 75+0.4 x=0.45 \cdot 75+0.45 x, 0.15 \cdot 75=0.05 x, x=225 \text {. }
$$
In total, it results in $75+225=300$ g of solution. | Answer: 300.
## Problem 9
Points $B_{1}$ and $C_{1}$ are the feet of the altitudes of triangle $ABC$, drawn from vertices $B$ and $C$ respectively. It is known that $AB=8, \quad AC=7, \quad \sin \angle BAC=\frac{3 \sqrt{5}}{7}$. Find the length of the segment $B_{1} C_{1}$.
## Solution:
Let the angle $\angle BAC$ b... | 300 | Algebra | math-word-problem | Incomplete | Yes | olympiads | false |
1. (10 points). Tourist Nikolai Petrovich was late by $\Delta t=5$ minutes for the departure of his riverboat, which had set off downstream. Fortunately, the owner of a fast motorboat agreed to help Nikolai Petrovich. Catching up with the riverboat and disembarking the unlucky tourist, the motorboat immediately set off... | 1. $s=\left(3 v_{\mathrm{T}}+v_{\mathrm{T}}\right) \cdot\left(t_{1}+\Delta t\right)=\left(5 v_{\mathrm{T}}+v_{\mathrm{T}}\right) \cdot t_{1} \rightarrow 4 \Delta t=2 t_{1} \rightarrow t_{1}=2 \Delta t=10$ min $t_{2}=\frac{s}{5 v_{\mathrm{T}}-v_{\mathrm{T}}}=\frac{6 v_{\mathrm{T}} \cdot t_{1}}{4 v_{\mathrm{T}}}=\frac{3}... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (30 points). To hang a New Year's decoration, Anya leans a ladder against a smooth wall so that its base is at the maximum possible distance from the wall. The coefficient of friction between the ladder and the floor is $\mu=2 / 3$, Anya's mass $M=70 \mathrm{kg}$, and the mass of the ladder $\mathrm{m}=20$ kg. The l... | 1.2 Two weights with masses \( m_{1}=180 \) g and \( m_{2}=120 \) g hang from the ends of a weightless, inextensible string, which is passed over a fixed pulley. Initially, the weights were at the same level. The weights were released and came into motion without initial velocity. What will be the vertical distance bet... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points). A thin-walled cylindrical glass, filled with maple syrup to $1 / 4$, floats in a container of water, submerging to the middle. The same glass, but filled with water to $1 / 2$, floats in a container of syrup, also submerging to the middle. What fraction of the glass can be filled with syrup so that it d... | 1.3 Equal volumes of two immiscible liquids with different densities were poured into a long thin tube, filling it halfway. The tube was then bent into a ring, placing it in a vertical plane (see fig.). The angle that the segment passing through the boundary of the liquids and the center of the ring makes with the vert... | \rho_{2}=\rho_{1}\cdot\frac{1-\tan\alpha}{1+\tan\alpha} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points). For the manufacture of a high-temperature vacuum furnace heater, 140 g of platinum wire was used. In the furnace, 400 g of silver in a light crucible was placed and heated from $t_{1}=20^{\circ} \mathrm{C}$ to $t_{2}=1000^{\circ} \mathrm{C}$ over $\tau=8$ minutes. To prevent the furnace body from heatin... | 1.4 An ideal monatomic gas is the working substance of a heat engine operating according to the thermodynamic cycle $1-2-3-1$. Find the efficiency of this heat engine if it is known that the maximum and minimum pressure values differ by a factor of $k=4$.
The efficiency by definition $\eta=\frac{A}{Q^{+}}$. The work c... | notfound | Algebra | math-word-problem | Yes | Incomplete | olympiads | false |
5. (10 points). Three small metal spheres, one with radius $R$ and two with radius $2R$, are charged with the same charge $+q$ and are located at the vertices of an equilateral triangle with a side length significantly exceeding the radii of the spheres. The spheres are connected by a wire, and then the wire is removed... | # 1.5 From eight identical segments of homogeneous wire of the same cross-section, a regular tetrahedral pyramid $A B C D F . \mathrm{K}$ was constructed. A source of constant voltage was connected to its vertices $A$ and $B$. Find the ratio of the thermal powers dissipated in the edges $A F$ and $C F$. By dividing the... | 16 | Other | math-word-problem | Yes | Incomplete | olympiads | false |
6. (15 points). A light bulb is hanging from the ceiling at a height of $\mathrm{H}=2$ m. Below it, glasses with an optical power of $D=+2$ diopters are placed horizontally so that clear images of the bulb are formed on the floor. The distance between the centers of the lenses in the glasses is $a=60$ mm. Find the dist... | 1.6 On the principal optical axis of a converging lens with a focal length of $F=50$ cm, two point light sources are located at equal distances $\ell=120$ cm to the left and right of it. How far should the lens be moved so that the images of the sources coincide?
 A small rubber ball moves between two massive vertical walls, colliding with them. One of the walls is stationary, while the other is moving away from it at a constant speed \( u = 100 \, \text{cm} / \text{s} \). Assuming the ball's motion is always horizontal and the collisions are perfectly elastic, fi... | 1. With each elastic collision with the moving wall, the direction of the ball's velocity changes to the opposite, and its speed relative to the Earth decreases by $2 u$, i.e., by 200 cm/s. Upon collision with the stationary wall, only the direction changes. Thus, after $n=10$ collisions with the moving wall and the sa... | 17\, | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) A satellite is launched vertically from the pole of the Earth at the first cosmic speed. To what maximum distance from the Earth's surface will the satellite travel? (The acceleration due to gravity at the Earth's surface $g=10 \mathrm{m} / \mathrm{c}^{2}$, radius of the Earth $R=6400$ km). | 2.
According to the law of conservation of energy: $\frac{m v_{I}^{2}}{2}-\frac{\gamma m M}{R}=-\frac{\gamma m M}{R+H}$
The first cosmic speed $v_{I}=\sqrt{g R}$
The acceleration due to gravity at the surface $g=\frac{\gamma M}{R^{2}}$
Then $\frac{m g R}{2}-m g R=-\frac{m g R^{2}}{R+H}$
Finally, $H=R=6400$ km.
(... | 6400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) Determine the mass $m$ of helium needed to fill an empty balloon of mass $m_{1}=10$ g so that the balloon will rise. Assume the temperature and pressure of the gas in the balloon are equal to the temperature and pressure of the air. The molar mass of helium $M_{\mathrm{r}}=4$ g/mol, the molar mass of air... | 3. For the balloon to take off, the condition is $F_{\text {Arch }}=\left(m+m_{1}\right) g$
$F_{\text {Arch }}=\rho_{0} g V$
From the Clapeyron-Mendeleev equation, the density of air is
$\rho_{0}=\frac{P_{0} M_{\mathrm{B}}}{R T_{0}}, \quad$ and the volume of the balloon $V=\frac{m}{M_{\mathrm{r}}} \frac{R T_{0}}{P_{... | 1.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (15 points) Carefully purified water can be supercooled to a temperature below $0^{\circ}$. However, if a crystal of ice is thrown into it, the water will immediately begin to freeze. What fraction of water supercooled to $-10^{\circ} \mathrm{C}$ in a thermos will freeze if a small ice chip is thrown into it? The sp... | 4. A small ice crystal becomes a center of crystallization. The heat released during crystallization goes to heating the water. Eventually, thermal equilibrium is established at $t=0^{\circ} \mathrm{C}$.
From the heat balance equation $\quad \lambda m_{\text{ice}}=c m_{\text{water}} \Delta t$
Finally $\quad \frac{m_{... | 0.13 | Other | math-word-problem | Yes | Yes | olympiads | false |
5. (25 points) Six resistors with resistances $R_{1}=1$ Ohm, $R_{2}=2$ Ohm, $R_{3}=3$ Ohm, $R_{4}=4$ Ohm, $R_{5}=5$ Ohm, and $R_{6}=6$ Ohm are connected in series and form a loop. A source of constant voltage is connected to the resulting circuit such that the resistance between its terminals is maximized. The voltage ... | 5. When connecting a source to any two contacts, the external circuit will represent two parallel branches. Then the total resistance of the circuit is
$R_{\text {total }}=\frac{R_{I} R_{I I}}{R_{I}+R_{I I}}$
For any method of connection, the sum of the resistances in the branches is the same (21 Ohms). The total res... | 4.32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (15 points) Looking down from the edge of the stream bank, Vovochka decided that the height of his rubber boots would be enough to cross the stream. However, after crossing, Vovochka got his legs wet up to his knees ($H=52$ cm). Estimate the height $h$ of Vovochka's boots. Assume the depth of the stream is constant,... | 6. From the figure
$\frac{d}{h}=\operatorname{tg} \alpha ; \quad \frac{d}{H}=\operatorname{tg} \beta ; \quad \frac{H}{h}=\frac{\operatorname{tg} \alpha}{\operatorname{tg} \beta}$
Since all the information about the bottom of the stream falls into the space limited by the eye's pupil, all angles are small: $\operatorn... | 39 | Other | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) The electric train "Sapsan", approaching a railway station at a speed of $v=216$ km/hour, gives a warning sound signal $\Delta t=5 \mathrm{s}$ long half a kilometer before it. What will be the duration of the signal $\Delta t_{1}$ from the perspective of passengers standing on the platform? The speed of ... | 1. The length of the "Sapsan" signal
$\lambda=(c-v) \Delta t$.
Duration of the signal from the passenger's perspective
$\Delta t_{1}=\frac{\lambda}{c}=\frac{(c-v)}{c} \Delta t=4.12 c$.
(10 points) | 4.12 | Other | math-word-problem | Yes | Problem not solved | olympiads | false |
2. (15 points) A wooden cube with edge $\ell=30$ cm floats in a lake. The density of wood $\quad \rho=750 \mathrm{kg} / \mathrm{m}^{3}, \quad$ the density of water $\rho_{0}=1000 \mathrm{kg} / \mathrm{m}^{3}$. What is the minimum work required to completely pull the cube out of the water? | 2. In equilibrium $m g=\rho_{0} g \frac{3}{4} \ell^{3}$
External minimal force at each moment in time
$$
F=m g-F_{\mathrm{Apx}}=\rho_{0} g \frac{3}{4} \ell^{3}-\rho_{0} g \ell^{2}\left(\frac{3}{4} \ell-x\right)=\rho_{0} g \ell^{2} x,
$$
where $x$ is the height of the lift at that moment. The external force depends l... | 22.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) The diagram shows the heating process of a constant mass of a monatomic ideal gas. Determine the efficiency of this process.
| 3. From the graph $P(T)$, it can be seen that $P \sim \sqrt{T} \rightarrow P^{2} \sim T$. From the Mendeleev-Clapeyron equation $P V=v R T$, therefore, in our case, $P \sim V$.
The work done by the gas is numerically equal to the area of the figure on the $P(V)$ diagram.
$$
A=\frac{1}{2}\left(2 P_{0}+P_{0}\right)\lef... | 0.25 | Other | math-word-problem | Incomplete | Yes | olympiads | false |
4. (15 points) At the vertices of a square with side $a$
there are four charges: $q, q, q, -q$. What are the magnitude of the electric field $E$ and the potential $\varphi$ at the midpoint of... | 4. Projections of the field intensity vector at the desired point
$E_{x}=2 E_{1}=\frac{8 k q}{a^{2}}$
$E_{y}=2 E_{4 y}=\frac{2 k q \cos \alpha}{r^{2}}$,
$r=\sqrt{a^{2}+\frac{a^{2}}{4}}=\frac{\sqrt{5}}{2} a ; \quad \cos \alpha=\frac{a}{r}=\frac{2}{\sqrt{5}}$.
Then $E_{y}=\frac{16 k q}{5 \sqrt{5} a^{2}}$
 A voltmeter connected to the terminals of a current source with an EMF of $12 \mathrm{~V}$ shows a voltage $U=9 \mathrm{~V}$. Another identical voltmeter is connected to the terminals of the source. Determine the readings of the voltmeters. (The internal resistance of the source is non-zero, the resistan... | 5. According to Ohm's law for a segment of a circuit and for a closed circuit
In the first case
$U_{1}=I_{1} R_{\mathrm{v}}=\frac{\varepsilon R_{\mathrm{v}}}{r+R_{\mathrm{v}}}=\frac{\varepsilon}{\left(\frac{r}{R_{\mathrm{v}}}+1\right)} \rightarrow \frac{r}{R_{\mathrm{v}}}=\frac{\varepsilon-U_{\mathrm{v}}}{U_{1}}$
In... | 7.2\mathrm{~V} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) A diverging lens produces an image of a nail that is reduced by a factor of 4. The nail is positioned on the principal optical axis of the lens with its head facing the lens. The length of the nail is $\ell=20$ cm, and the optical power of the lens is $D=-5$ diopters. Find the distance from the optical c... | 6. According to the problem:
$F=\frac{1}{D}=20 \mathrm{~cm}=\ell$
$d_{1}=d_{2}-F ; \quad f_{2}-f_{1}=\frac{1}{4} F$
From the thin lens formulas for the two ends of the nail, we find the distances to the ends of the image
$-\frac{1}{F}=\frac{1}{d_{2}-F}-\frac{1}{f_{1}} \rightarrow f_{1}=\frac{F\left(d_{2}-F\right)}{... | 11.2 | Other | math-word-problem | Yes | Yes | olympiads | false |
1. Dmitry is three times as old as Grigory was when Dmitry was as old as Grigory is now. When Grigory becomes as old as Dmitry is now, the sum of their ages will be 49 years. How old is Grigory? | Solution: Let Gregory be $y$ years old in the past, and Dmitry be $x$ years old. Then currently, Gregory is $x$ years old, and Dmitry is $3 y$ years old. In the future, Gregory will be $3 y$ years old, and Dmitry will be $z$ years old, and according to the condition, $z+3 y=49$. Since $z-3 y=3 y-x ; 3 y-x=x-y$, then $9... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the rational number - the value of the expression
$$
2 \cos ^{6}(5 \pi / 16)+2 \sin ^{6}(11 \pi / 16)+3 \sqrt{2} / 8
$$ | Solution: By the reduction formula, the desired expression is equal to
$2 \cos ^{6}(5 \pi / 16)+2 \sin ^{6}(5 \pi / 16)+3 \sqrt{2} / 8$. By the sum of cubes formula, we have:
$\cos ^{6}(5 \pi / 16)+\sin ^{6}(5 \pi / 16)=1-\frac{3 \sin ^{2}(10 \pi / 16)}{4}$. Therefore, our expression takes the form:
$2-\frac{3 \sin ... | \frac{5}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $\frac{4}{\sqrt{\log _{3}(81 x)}+\sqrt{\log _{3} x}}+\sqrt{\log _{3} x}=3$. | Solution: Using the properties of logarithms, our equation can be rewritten as $\frac{4}{\sqrt{4+\log _{3} x}+\sqrt{\log _{3} x}}+\sqrt{\log _{3} x}=3$. Let $t=\log _{3} x$. Then $\frac{4}{\sqrt{4+t}+\sqrt{t}}+\sqrt{t}=3$. Multiplying the numerator and denominator of the first fraction by $\sqrt{t+4}-\sqrt{t}$, we arri... | 243 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the minimum value of the function $y=\sin 2x-(\sin x+\cos x)+1$. | Solution: Let $t=\sin x+\cos x$. Then $t^{2}=1+\sin 2 x$ and $y=t^{2}-t$.
In this case, $t \in[-\sqrt{2} ; \sqrt{2}]$. This function reaches its minimum at $t=1 / 2$ and this minimum is equal to $-1 / 4$. Answer: $-1 / 4$. | -\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The sum of $n$ terms of the geometric progression $\left\{b_{n}\right\}$ is 6 times less than the sum of their reciprocals. Find the product $b_{1} b_{n}$. | Solution: Let $S_{n}=b_{1}+b_{2}+\ldots+b_{n}, S_{n}^{1}=\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n}} \cdot$ Then
$b_{1} b_{n} S_{n}^{1}=\frac{b_{1} b_{n}}{b_{1}}+\frac{b_{1} b_{n}}{b_{2}}+\ldots+\frac{b_{1} b_{n}}{b_{n}}=b_{n}+b_{n-1}+\ldots+b_{1}=S_{n}$. From this we obtain: $b_{1} b_{n}=\frac{S_{n}}{S_{n}... | \frac{1}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Solve the system of equations $\left\{\begin{array}{c}x^{4}-y^{4}=3 \sqrt{|y|}-3 \sqrt{|x|} \\ x^{2}-2 x y=27\end{array}\right.$. | Solution: Write the system as $\left\{\begin{array}{c}x^{4}+3 \sqrt{|x|}=y^{4}+3 \sqrt{|y|} \\ x^{2}-2 x y=27\end{array}\right.$. The function $z=s^{4}+3 \sqrt{|s|}$ is even and, moreover, it is increasing on $[0 ;+\infty)$. Therefore, from the first equation, we find that $x=y$ or $x=-y$. Substituting $x=y$ into the s... | \(3,-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. On the sides $B C, C A, A B$ of an equilateral triangle $A B C$ with side length 7, points $A_{1}, B_{1}, C_{1}$ are taken respectively. It is known that $A C_{1}=B A_{1}=C B_{1}=3$. Find the ratio of the area of triangle $A B C$ to the area of the triangle formed by the lines $A A_{1}, B B_{1}, C C_{1}$. | Solution:
Triangles $A B A_{1}, B C B_{1}, C A C_{1}$ are equal by sides and the angle between them, triangles $A D C_{1}, B E A_{1}, C F B_{1}$ are equal by side and angles. Triangle $A D C_{1}$ is similar to $A B A_{1}$ by two angles, triangle $A B C$ is similar to $D E F$. Let $S=S_{A B C}, S_{1}=S_{A B A_{1}}, S_{... | 37 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. For what values of the parameter $a$ does the equation $x^{4}-20 x^{2}+64=a\left(x^{2}+6 x+8\right)$ have exactly three distinct solutions? | Solution: Factorize the right and left sides of the equation:
\((x-2)(x+2)(x-4)(x+4)=a(x+2)(x+4)\). The equation can be written as \((x+2)(x+4)\left(x^{2}-6 x+8-a\right)=0\). It is obvious that -2 and -4 are roots of this equation. We are satisfied with the situation where exactly one of the roots of the equation \(x^... | -1;24;48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Ivan is twice as old as Peter was when Ivan was as old as Peter is now. When Peter becomes as old as Ivan is now, the sum of their ages will be 54 years. How old is Peter? | Solution: Let Peter's age in the past be $y$ years, and Ivan's age be $x$ years. Then currently, Peter is $x$ years old, and Ivan is $2 y$ years old. In the future, Peter will be $2 y$ years old, and Ivan will be $z$ years old, and according to the condition, $z+2 y=54$. Since $z-2 y=2 y-x ; 2 y-x=x-y$, then $6 y-x=54$... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the rational number - the value of the expression
$$
\cos ^{6}(3 \pi / 16)+\cos ^{6}(11 \pi / 16)+3 \sqrt{2} / 16
$$ | Solution: By the reduction formula, the desired expression is equal to $\sin ^{6}(5 \pi / 16)+\cos ^{6}(5 \pi / 16)+3 \sqrt{2} / 16$. By the sum of cubes formula, we have: $\cos ^{6}(5 \pi / 16)+\sin ^{6}(5 \pi / 16)=1-\frac{3 \sin ^{2}(10 \pi / 16)}{4}$. Therefore, our expression takes the form: $1-\frac{3 \sin ^{2}(1... | \frac{5}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $\frac{1}{\sqrt{\log _{5}(5 x)}+\sqrt{\log _{5} x}}+\sqrt{\log _{5} x}=2$. | Solution: Using the properties of logarithms, our equation can be rewritten as $\frac{1}{\sqrt{1+\log _{5} x}+\sqrt{\log _{5} x}}+\sqrt{\log _{5} x}=2$. Let $t=\log _{5} x$. Then $\frac{1}{\sqrt{1+t}+\sqrt{t}}+\sqrt{t}=2$. By multiplying the numerator and denominator of the first fraction by $\sqrt{t+1}-\sqrt{t}$, we a... | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The sum of $n$ terms of the geometric progression $\left\{b_{n}\right\}$ is 8 times less than the sum of their reciprocals. Find the product $b_{1} b_{n}$. | Solution: Let $S_{n}=b_{1}+b_{2}+\ldots+b_{n}, S_{n}^{1}=\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n}}$. Then $b_{1} b_{n} S_{n}^{1}=\frac{b_{1} b_{n}}{b_{1}}+\frac{b_{1} b_{n}}{b_{2}}+\ldots+\frac{b_{1} b_{n}}{b_{n}}=b_{n}+b_{n-1}+\ldots+b_{1}=S_{n}$. From this we obtain: $b_{1} b_{n}=\frac{S_{n}}{S_{n}^{1}}=... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. On the sides $B C, C A, A B$ of an equilateral triangle $A B C$ with side length 11, points $A_{1}, B_{1}, C_{1}$ are taken respectively. It is known that $A C_{1}=B A_{1}=C B_{1}=5$. Find the ratio of the area of triangle $A B C$ to the area of the triangle formed by the lines $A A_{1}, B B_{1}, C C_{1}$. | Solution:
Triangles $A B A_{1}, B C B_{1}, C A C_{1}$ are equal by sides and the angle between them, triangles $A D C_{1}, B E A_{1}, C F B_{1}$ are equal by side and angles. Triangle $A D C_{1}$ is similar to $A B A_{1}$ by two angles, triangle $A B C$ is similar to $D E F$. Let $S=S_{A B C}, S_{1}=S_{A B A_{1}}, S_{... | 91 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. At the base of the quadrilateral pyramid $S A B C D$ lies the quadrilateral $A B C D$, in which $A B=A D=6, C B=C D=2$, and sides $A B$ and $B C$ are mutually perpendicular. All lateral faces are inclined to the base plane at an angle of $60^{\circ}$. Find the volume $V$ of the pyramid, knowing that $V>12$. | Solution: At the base of the pyramid lies a quadrilateral composed of two equal right-angled triangles. The area of the base is $S=6 \cdot 2=12$. Since the lateral faces are inclined to the base plane at the same angle, the apex of the pyramid projects onto the base plane at a point equidistant from the lines $A B, B C... | 12\sqrt{3} | Geometry | math-word-problem | Yes | Incomplete | olympiads | false |
10. For what values of the parameter $a$ does the equation $x^{4}-40 x^{2}+144=a\left(x^{2}+4 x-12\right)$ have exactly three distinct solutions? | Solution: Factorize the right and left sides of the equation:
$(x-2)(x+2)(x-6)(x+6)=a(x-2)(x+6)$. The equation can be written as
$(x-2)(x+6)\left(x^{2}-4 x-12-a\right)=0$. It is obvious that 2 and -6 are roots of this equation. We are satisfied with the situation where exactly one of the roots of the equation $x^{2}-4... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Is it possible to color each $1 \times 1$ cell of a $1000 \times 1000$ square in one of three colors: red, blue, and green, such that each red cell has two adjacent blue cells, each blue cell has two adjacent green cells, and each green cell has two adjacent red cells? Cells are considered adjacent if they share a c... | 5.1
## Solution.
Suppose such a coloring exists, and the top-left cell is red. Then the two adjacent cells are blue (see the figure). But then the next three cells, forming a diagonal, are green. The next diagonal consists of red cells, and so on. But then both neighbors of the red cell in the opposite corner are gre... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 6.1
From the minimality, it follows that $n-1=m^{2}$ and, by the condition, $n+2015<(m+1)^{2}$. Therefore, $2016<(m+1)^{2}-m^{2}=2 m+1$. From this, $m \geq 1008$. It remains to note that $n=1008^{2}+1$ fits.
№13
Since the coefficient of $x^{2}$ is positive for any value of the parameter $a$, the condition of the p... | Answer. $a \in(-2-\sqrt{11} ;-2+\sqrt{11})$.
#14
Subtracting the second equation from the first, we get: $\sin \alpha-\cos \alpha-\cos (\beta+\gamma)=\frac{1}{2}$, which means $\sin \alpha-\cos \alpha-\cos (\pi-\alpha)=\sin \alpha=\frac{1}{2}$. There are two possible cases:
1) $\alpha=30^{\circ}$. Then $\sin \beta \... | 1008^{2}+1,\in(-2-\sqrt{11};-2+\sqrt{11}),30,30,120,15,75,90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. The braking distance of a certain body moving on a rough surface turned out to be 60 m. Find the speed of the body before the start of the movement. The coefficient of friction is known and is equal to 0.5. | 13.1
№19
Solution: The first configuration - the lateral sides of an isosceles triangle connect the source and its images. The second - the lateral sides of an isosceles triangle - one connects the images, and the other connects the source and one of the images. In the first case, the angle between the directions fro... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 12. Task 12
In the insurance contract, the insurance amount is set at 500,000 rubles and a conditional franchise of $1 \%$. The actual damage amounted to 4000 rubles. What will be the amount of insurance compensation? | Write the answer as a number without spaces, without units of measurement and any signs.
# | 3960 | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
# 13. Problem 13
Maxim deposited 1000 rubles in a bank for a term of 1 year at an annual interest rate of $12 \%$ with monthly capitalization of interest. After one month, he closed his bank deposit. How much money will Maxim receive? | Write the answer as a number without spaces, without units of measurement and any signs.
# | 1010 | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
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