problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
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 | 2,552 |
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 | 2,554 |
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 | 2,556 |
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 | 2,558 |
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 | 2,559 |
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 | 2,560 |
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 | 2,561 |
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 | 2,566 | |
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 | 2,567 |
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 | 2,568 |
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 | 2,572 |
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 | 2,573 | |
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 | 2,577 |
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 | 2,578 |
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 | 2,582 |
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 | 2,585 |
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 | 2,587 |
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 | 2,588 |
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 | 2,589 |
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 | 2,591 |
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 | 2,593 |
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 | 2,594 |
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 | 2,595 |
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 | 2,596 |
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 | 2,597 |
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 | 2,599 |
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 | 2,605 | |
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 | 2,607 |
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 | 2,608 |
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 | 2,609 | |
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 | 2,610 |
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
$$
There are two possible cases:
Case 1. $\left\{\begin{array}{l}{\left[2 x-x^{2}\right]=-1} ... | x_{\}=-\frac{1}{3},x_{\max}=\frac{7}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,611 |
3. For which $a$ does the system $\left\{\begin{array}{l}x \cos a+y \sin a=5 \cos a+2 \sin a \\ -3 \leq x+2 y \leq 7,-9 \leq 3 x-4 y \leq 1\end{array}\right.$ have a unique solution? | Solution. The inequalities in the system limit a parallelogram:
$$
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 | 2,612 |
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 $x-1$:
$$
x^{2}-5(n-2) x+6\le... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,613 |
5. The midline of a trapezoid is 4. A line parallel to the bases of the trapezoid and dividing its area in half intersects the lateral sides at points $M$ and $N$. Find the smallest possible length of the segment $M N$. | # Solution.
Notations: $A D=a, B C=b, C K=H, N E=h$ - heights of triangles $N C P$ and $D N Q$, $M N=x$
$$
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 | 2,614 |
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 | 2,617 |
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 | 2,619 |
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 | 2,621 |
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 | 2,622 | |
5. Can a regular triangle be divided into 2025 equal regular triangles? Justify your answer.
## Answers and solutions
Problem 1 Answer: $54^{0}$
Problem 2 Answer: 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 | 2,624 |
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:
=\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 | 2,632 |
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 | 2,637 |
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 | 2,638 |
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 | 2,639 |
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 | 2,640 |
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 | 2,641 |
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
$$
V=\frac{1}{3} S_{S M N} \cdot h=\frac{1}{6} m n \sin \alpha \cdot h=\frac{1}{6} m n... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,642 |
1. Let $\{x\}$ and $[x]$ denote the fractional and integer parts of the number $x$. The integer part of the number $x$ is the greatest integer not exceeding $x$, and $\{x\}=x-[x]$. Find $x$ for which $4 x^{2}-5[x]+8\{x\}=19$. | Solution.
$$
\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}
$$
If $x$ is the desired value, then the quadratic trinomial $f(t)=4 t^{2}+8([x]+1) t+4[x]^{2}-5[x]-19$ has a root $t=\{x\}$ on the interval $[0 ; 1)$.
Cas... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,643 |
2. Find the integers $n$ for which the expression $\frac{1}{12}\left(8 \sin \frac{\pi n}{10}-\sin \frac{3 \pi n}{10}+4 \cos \frac{\pi n}{5}+1\right)$ takes integer values. | Solution. Let's introduce the notation $t=\sin \frac{\pi n}{10} \in[-1 ; 1]$ and rewrite the original expression:
$$
\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 | 2,644 |
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 | 2,645 |
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
$$
\left\{\begin{array} { l }
{ f ( - 1 ) \leq 0 } \\
{ f ( - 2 ) \leq 0 }
\end{array} \rightarrow \left\{\begin{array} { c }
{ 1 - ( 2 \xi + 1 ) + 3 - \xi \leq 0 } \\
{ 1 - 2 ( 2 \xi + 1 ) + 3 - \xi \leq 0 }
\end{arr... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,646 |
5. For what values of $a$ does the system $\left\{\begin{array}{c}(x-7 \cos a)^{2}+(y-7 \sin a)^{2}=1 \\ |x|+|y|=8\end{array}\right.$ have a unique solution? | Solution. The set of points on the plane, with coordinates $(x ; y)$ that satisfy the first equation of the system, lie on a circle $K_{a}$ of radius 1 centered at the point $O(7 \cos a ; 7 \sin a)$. As $a$ changes, the center moves along a circle of radius 7 centered at the origin.
\pi}{4},k\inZ | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,647 |
6. In a triangular pyramid $S A B C$, the angle $A S B$ at vertex $S$ is $30^{\circ}$, and the lateral edge $S C$ is inclined to the plane of the face $A S B$ at an angle of $45^{\circ}$. The sum of the lengths of the lateral edges of the pyramid is 9. Find the greatest possible value of the volume of the pyramid under... | # Solution.
Let's introduce the notations: $S A=a, S B=b, S C=c, C K=h, \square A S B=\alpha, \square C S K=\beta$,
$C K$ is the perpendicular to the plane $A S B$. Then
$$
V_{S A B C}=\fr... | \frac{9\sqrt{2}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,648 |
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,
$$
y_{2... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 2,649 |
1. There are 12 carriages in the train, and they contain 384 passengers. In the second carriage, there are more than two passengers more than in the first, in the third carriage, there are more than two passengers more than in the second, and so on until the last carriage. The number of passengers in the last carriage ... | Answer: There are no solutions.
Solution. Let $y_{k}$ be the number of passengers in the carriage with number $k, k=1,2,3, \ldots, 12$. According to the problem, $\sum_{k=1}^{12} y_{k}=384$. Additionally, it is stated that $y_{2} \geq y_{1}+3, y_{3} \geq y_{1}+6, \ldots, y_{11} \geq y_{1}+30, y_{12} \geq y_{1}+33$. Ad... | nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,650 |
2. Find the largest solution of the equation on the interval $(0 ; 2 \pi)$
$$
(\sin x + \cos x + \sin 3x)^{3} = \sin^{3} x + \cos^{3} x + \sin^{3} 3x
$$ | Answer: $x=\frac{15 \pi}{8}$. | \frac{15\pi}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,651 |
2. Find the largest solution of the equation on the interval $(0 ; 2 \pi)$
$$
(\cos 2 x+\sin 3 x+\cos 4 x)^{3}=\cos ^{3} 2 x+\sin ^{3} 3 x+\cos ^{3} 4 x
$$ | Answer: $x=\frac{25 \pi}{14}$. | \frac{25\pi}{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,654 |
3. Find the first 2000 digits after the decimal point in the decimal representation of the number $(9+4 \sqrt{5})^{2021}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Answer: all 2000 digits after the decimal point are nines. | 999\ldots999 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 2,655 |
2. Find the largest solution of the equation on the interval $(0 ; 2 \pi)$
$$
(\cos 3 x+\cos 4 x+\cos 5 x)^{3}=\cos ^{3} 3 x+\cos ^{3} 4 x+\cos ^{3} 5 x
$$ | Answer: $x=\frac{17 \pi}{9}$. | \frac{17\pi}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,657 |
4. Solve the equation $\left|5 x-\sqrt{1-25 x^{2}}\right|=5 \sqrt{2} x\left(100 x^{2}-3\right)$. | Answer: $x=\frac{\sqrt{2+\sqrt{2}}}{10} ; x=-\frac{\sqrt{2-\sqrt{2}}}{10}$. | \frac{\sqrt{2+\sqrt{2}}}{10};-\frac{\sqrt{2-\sqrt{2}}}{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,659 |
0.35 \cdot 160+0.1 x=0.2 \cdot 160+0.2 x, 0.15 \cdot 160=0.1 x, x=240 \text {. }
$$
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 | 2,660 |
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 | 2,662 |
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 | 2,663 |
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 | 2,664 |
1. (15 points) 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 | 2,668 |
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 | 2,669 |
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 | 2,670 |
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 | 2,671 |
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 | 2,672 |
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 | 2,673 |
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 | 2,675 |
5. (20 points) 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 | 2,678 |
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 | 2,679 |
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 | 2,680 |
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 | 2,681 |
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 | 2,682 |
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 | 2,683 |
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 | 2,684 |
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 | 2,685 |
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 | 2,686 |
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 | 2,687 |
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 | 2,688 |
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 | 2,689 |
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 | 2,690 |
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 | 2,691 |
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 | 2,692 |
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 | 2,694 |
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 | 2,695 |
# 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 | 2,696 |
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 | 2,697 |
# 16. Problem 16
Calculate the land tax for a plot of 15 acres if the cadastral cost of one acre is 100000 rubles, and the tax rate is $0.3 \%$. | Write the answer as a number without spaces, without units of measurement and any signs.
# | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,702 | |
# 18. Problem 18
The personal income tax is $13 \%$ of the amount of income accrued to the employee. If the employee wants to receive 20000 rubles in hand (accrued income minus personal income tax), determine the amount of accrued income. | Write the answer as a number without spaces, without units of measurement and any signs, without cents.
# | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,704 | |
01.04.22 Fedor opened a deposit in the bank for 3 months with the possibility of replenishing funds and placed on it:
- the amount of dividends received from company $\mathrm{ABC}$;
- the amount of the income tax refund for 2021 from the tax office;
- an amount of 4100 rubles.
The annual interest rate on the deposit ... | # Solution
1) Specify the amount and name of the tax deductions that each spouse will receive for 2021. Explain your answer.
Fedor:
The investment tax deduction is limited to: 400,000 rubles (Fedor's annual salary was 720,000 rubles, so he is entitled to an investment tax deduction of 400,000 rubles).
Social tax de... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,709 |
From channel A to the Wiki site, $850 * 0.06=51$ people will transition
From channel B to the Wiki site, $1500 * 0.042=63$ people will transition
From channel C to the Wiki site, $4536 / 72=63$ people will transition | Answer: The most people will transition from channels B and V - 63 people each
## 2
Cost of transition from advertising on channel A: $-3417 / 51 = 67$ rubles
Cost of transition from advertising on channel B: $4914 / 63 = 78$ rubles
Answer: The lowest cost of transition to the site from advertising on channel A - 6... | 2964 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,711 |
5.
Expenses for medical services provided to a taxpayer's child (under 18 years of age) by medical organizations
## Question 11
Score: 6.00
An investor has a brokerage account with an investment company. In 2021, the investor received the following income from securities:
- dividends from shares of JSC “Winning” ... | Answer and write it in rubles as an integer without spaces and units of measurement.
Answer: $\qquad$
Question 12
Score: 5.00
Insert the missing terms from the drop-down list.
Under the insurance contract, one party
insured; insurance premium; beneficiary; insurer; insurance amount
undertakes for the fee stipulat... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,713 |
5.
The higher the risk of a financial instrument, the higher its return
Question 15
Score: 6.00
Agnia's monthly salary in 2021 was 60,000 rubles (before taxation). It is known that on 01.01.2021, Agnia opened the following deposits in banks (she had no deposits before):
| Bank | Deposit amount, rub. | Interest ra... | Answer write in rubles as an integer without spaces and units of measurement.
Answer:
Question 16
Score: 5.00
Establish the correspondence between specific taxes and their types.
| personal income tax | federal tax; local tax; regional tax; |
| :---: | :---: |
| land tax | federal tax; local tax; regional tax; |... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,714 |
# 14. Problem 14
Finding himself in a difficult financial situation, Ivan Karlavich Buratino, unable to find a better solution, decided to seek financial assistance from the notorious criminal authority, Basilio, also known as Cat. Cat offered Buratino a loan of 100 coins for a period of 128 days under two alternative... | Round the answer to the nearest whole number and write it without the unit of measurement. In case of overpayment, put a minus sign before the number, in case of savings, do not put a sign.
# | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,716 |
# 17. Problem 17
A photographer at a maternity hospital photographs all families who come to take the mother and baby home. A few days later, after making a set of photographs, the photographer offers the families to buy the photographs at a price of 600 rubles per set. The photographer's expenses amount to 100 rubles... | Give the answer in percentages, but without the "%" sign. For example, if your answer is 15%, write only 15 in your response.
# | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,718 |
# 20. Problem 20
A commercial bank offers the following interest accrual scheme (simple interest) for a term deposit with a duration of one year:
- for the first 3 months, interest is accrued at a rate of $12 \%$ per annum;
- for the next 3 months, interest is accrued at a rate of $8 \%$ per annum;
- for all the rema... | Give the answer in percentages, but without the "%" sign. For example, if your answer is 15%, write only 15 in your answer. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 2,719 |
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