problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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3. On the sides $B C$ and $C D$ of the square $A B C D$, points $E$ and $F$ are chosen such that the angle $E A F$ is $45^{\circ}$. The length of the side of the square is 1. Find the perimeter of triangle $C E F$. Justify your solution. | # Solution:
If point $D$ is reflected relative to line $A F$, and then relative to line $A E$, it will transition to point $B$. Indeed, the composition of two axial symmetries relative to intersecting lines is a rotation by twice the angle between the lines. In our case, these two symmetries are equivalent to a $90^{\... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,644 |
4. It is known that the number $\cos 6^{0}$ is a root of the equation $32 t^{5}-40 t^{3}+10 t-\sqrt{3}=0$. Find the other four roots of this equation. (Answers in the problem should be compact expressions, not containing summation signs, ellipses, and radicals.) | Solution:
$\cos 5 x=16 \cos ^{5} x-20 \cos ^{3} x+5 \cos x$. Then the other roots of the equation will be the numbers $\cos 66^{\circ}, \cos 78^{\circ}, \cos 138^{\circ}, \cos 150^{\circ}$.
Answer: $\cos 66^{\circ}, \cos 78^{\circ}, \cos 138^{\circ}, \cos 150^{\circ}$ | \cos66,\cos78,\cos138,\cos150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,645 |
5. Find the area of the polygon, the coordinates of the vertices of which are the solutions of the system of equations $\left\{\begin{array}{l}x^{4}+\frac{7}{2} x^{2} y+2 y^{3}=0 \\ 4 x^{2}+7 x y+2 y^{3}=0\end{array}\right.$. | # Solution:
The system of equations has only three real solutions: $(0 ; 0),(2 ;-1),\left(-\frac{11}{2} ;-\frac{11}{2}\right)$. The area of the triangle with vertices at these coordinates is $16 \frac{1}{2}$.
Answer: $16 \frac{1}{2}$ | 16\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,646 |
6. Points $F$ and $G$ are chosen on the sides $AB$ and $BC$ of rectangle $ABCD$, respectively. A perpendicular $FK$ is dropped from point $F$ to side $CD$. A perpendicular $GH$ is dropped from point $G$ to side $AD$. The intersection point of $FK$ and $GH$ is denoted as $E$. Find the area of triangle $DFG$, given that ... | # Solution:
Let $A D=a, D C=b, H D=x$, and $D K=y$.
$$
\begin{aligned}
S_{D F G}=S_{A B C D} & -S_{A F D}-S_{F G B}-S_{D G C}=a b-\frac{1}{2} a y-\frac{1}{2}(b-y)(a-x)-\frac{1}{2} b x \\
&... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,647 |
1. (2 points) Find all numbers $x, y$ for which the equality
$$
4 x^{2}+4 x y \sqrt{7}+8 y^{2}+2 y \sqrt{3}-\frac{2 y}{\sqrt{3}}+\frac{7}{3}=1
$$ | 1. $y=-\frac{2}{\sqrt{3}}, \quad x=\frac{\sqrt{7}}{\sqrt{3}}$.
+ (2 points) - solution is correct
- (0 points) - there are errors in the solution, including arithmetic errors | -\frac{2}{\sqrt{3}},\quad\frac{\sqrt{7}}{\sqrt{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,648 |
2. (3 points) Let two sequences of numbers $\left(x_{0}, x_{1}, \ldots, x_{2009}\right)$, $\left(y_{0}, y_{1}, \ldots, y_{2009}\right)$ be constructed according to the following rules:
a) $x_{0}=12, x_{1}=\frac{1}{3}, y_{0}=4, y_{1}=\frac{1}{18}$
b) $x_{i+1}=x_{i-1}+4 x_{i}$ and $y_{i+1}=y_{i-1}-4 y_{i}$ for $i=1, \ld... | 2. 2 .
$+(3$ points) - the solution is correct
+- (2 points) - the idea of the recurrence relation is proven, but there are arithmetic errors in the solution
-+ (1 point) - the idea of the recurrence relation is formulated, but the problem is not completed | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,649 |
3. (3 points) Find all solutions of the inequality $\cos 5+2 x+x^{2}<0$, lying in the interval $\left[-2 ;-\frac{37}{125}\right]$. | 3. $x \in\left(-1-\sqrt{1-\cos 5} ;-\frac{37}{125}\right]$
+(3 points) - solution is correct
+- (2 points) - solution is correct, but in the process of comparing $-1+\sqrt{1-\cos 5}$ with $-\frac{37}{125}$
not all steps are sufficiently justified (for example, it is not explained why $\left.\cos 5<\frac{37}{125} \cd... | x\in(-1-\sqrt{1-\cos5};-\frac{37}{125}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,650 |
4. (4 points) In a parallelogram with sides 4 and 7, the bisectors of the four interior angles are drawn. Find the ratio of the area of the quadrilateral formed by the intersection of the bisectors to the area of the parallelogram. | 4. $\frac{S_{Q R T F}}{S_{A B C D}}=\frac{9}{56}$.
+ (4 points) - solution is correct.
+- (3 points) - solution is correct, but there are minor inaccuracies.
-+ (2 points) - solution is correct, but there are significant gaps (for example, it is not justified that the angle bisectors form a rectangle). | \frac{9}{56} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,651 |
5. (5 points) Find all solutions of the system of equations $\left\{\begin{array}{l}\frac{y^{3}}{x^{2}}+\frac{x}{y}=1 \\ \frac{2 x}{y^{2}}+\frac{4 y}{x}=1\end{array}\right.$. | 5. $x_{1}=-8, \quad y_{1}=-4, \quad x_{2}=-54, \quad y_{2}=-18$
+ (5 points) - solution is correct.
+- (3 points) - solution is generally correct, but there are unfortunate arithmetic errors.
-+ (2 points) - only one of the two solutions is found in the solution.
6. incorrect. | x_{1}=-8,\quady_{1}=-4,\quadx_{2}=-54,\quady_{2}=-18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,652 |
1. A 100-digit number has the form $a=1777 \ldots 76$ (with 98 sevens in the middle). The number $\frac{1}{a}$ is represented as an infinite periodic decimal. Find its period. Justify your answer. | # Solution:
Notice that $a=16 \cdot 111 \ldots 11$. The last number $b$ consists of 99 ones. According to the rules for converting a common fraction to a decimal, the number $\frac{1}{b}=0,(00 \ldots 09)$. Its period is 99. Then, when multiplying this fraction by the number $\frac{1}{16}=0.0625$, the period will not c... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,654 |
2. There are 5 cells. Two players take turns writing one of the digits 1 or 2 from left to right in these cells. If the resulting 5-digit number is divisible by 3, the player who made the first move wins; if it is not divisible by 3, the second player wins. Which player will win with correct play and any play by the op... | # Solution:
Let's recall the divisibility rule for 3: an integer is divisible by 3 if and only if the sum of all its digits is divisible by 3.
We will outline a winning strategy for the second player. On each turn, if the first player writes 1, the second player writes 2. If the first player writes 2, the second play... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,655 |
3. Let $A$ be the set of all integers that can be represented in the form $x^{2}+2 y^{2}$ where $x, y$ are integers. Let $B$ be the set of all integers that can be represented in the form $x^{2}$ $6 x y+11 y^{2}$ where $x, y$ are integers (for example, $6 \in A$, since $6=2^{2}+2 \cdot 1^{2}$ ). Are the sets $A$ and $B... | # Solution:
Rewrite the elements of set $B$ as $x^{2}-6 x y+11 y^{2}=(x-3 y)^{2}+2 y^{2}$. From this, it is clear that any element of set $A$ is an element of set $B$ and vice versa. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,656 |
4. The bases of trapezoid $A B C D$ are related by $A D=4 \cdot B C$, and the sum of angles $\angle A+\angle D=120^{\circ}$. Points $M$ and $N$ are chosen on the lateral sides such that $C N: N D=B M: M A=1: 2$. Perpendiculars drawn from points $M$ and $N$ to the lateral sides of the trapezoid intersect at point $O$. F... | # Solution:
Extend the lateral sides until they intersect at point $S$. Angle $S$ is $60^{\circ}$. Triangles $A S D$ and $B S C$ are similar; the similarity coefficient is 4. Moreover, by the condition, $A M$ is twice as long as $B M$. From this, it is easy to notice that $B S = B M$. Therefore, $A M = M S$. Thus, $M ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,657 |
5. Let $a_{n, m}$ denote the number obtained by writing down all numbers from $n$ to $m$ consecutively, where $n$ and $m$ are natural numbers, and $n > m \geq 1$. For example, the number $a_{4,2}=432$, and the number $a_{11,7}=1110987$. Prove that among such numbers, there is a number divisible by 2021. | # Solution:
Consider numbers of the form $a_{n, 1}$. Since there are infinitely many such numbers, there will be two numbers $a_{n, 1}$ and $a_{k, 1}, n>k$, that have the same remainder when divided by 2021. Then the difference $a_{n, 1}-a_{k, 1}$ is divisible by 2021. At the same time, $a_{n, 1}-a_{k, 1}=a_{n, k+1} \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,658 |
6. Prove that the system of equations has no solutions.
$$
\left\{\begin{array}{c}
x^{3}+x+y+1=0 \\
y x^{2}+x+y=0 \\
y^{2}+y-x^{2}+1=0
\end{array}\right.
$$ | # Solution:
Notice that $y\left(x^{3}+x+y+1\right)-x\left(y x^{2}+x+y\right)=y^{2}+y-x^{2}=0$, which contradicts the third equation of the system. | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,659 |
7. Let's fix 10 natural numbers $n_{1}, n_{2}, \ldots, n_{10}$ and denote their sum by $n=n_{1}+\cdots+n_{10}$. Suppose now that on a board, $n$ numbers $a_{1}, \ldots, a_{n}$ are written in a row, each of which is either 0 or 1. These numbers (in the order they are written) are divided into 10 groups:
$$
\underbrace{... | # Solution:
The desired number of sets is found by summing the number of sets with a given number of $\mathrm{k}$ non-zero groups:
For $\mathrm{k}=0$, such a set is unique;
For $\mathrm{k}=2 \quad \sum_{1 \leq i<j \leq 10}\left(2^{n_{i}}-2\right) \cdot\left(2^{n_{j}}-2\right)$
for $\mathrm{k}=4 \sum_{1 \leq i<j<l<s... | 2^{n-1}+\frac{1}{2}\cdot(2^{n_{1}}-2)\cdot(2^{n_{2}}-2)\cdot\ldots\cdot(2^{n_{10}}-2) | Combinatorics | proof | Yes | Yes | olympiads | false | 8,660 |
1. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent ... | Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be t... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,662 |
2. Real numbers $x, y, z$ satisfy the relations:
$$
4 x^{2}-2 x-30 y z=25 y^{2}+5 y+12 x z=9 z^{2}-3 z-20 x y .
$$
Find the maximum of the sum $a+b+c$, where $a=2 x+5 y, b=3 z+5 y, c=3 z-2 x$. | Solution. Note that
$$
a-b+c=0
$$
Let $A=4 x^{2}-2 x-30 y z, B=25 y^{2}+5 y+12 x z$ and $C=9 z^{2}-3 z-20 x y$. Subtracting these equations from each other, we get
$$
\begin{aligned}
& A-B=a \cdot(2 x-6 z-5 y-1)=0 \\
& B-C=b \cdot(5 y+4 x-3 z+1)=0 \\
& A-C=c \cdot(1-2 x-10 y-3 z)=0
\end{aligned}
$$
Assume that all ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,663 |
3. Find the value of $f(2019)$, given that $f(x)$ simultaneously satisfies the following three conditions:
1) $f(x)>0$ for any $x>0$
2) $f(1)=1$;
3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}$ for any $a, b \in \mathbb{R}$. | Solution. In the identity given in the problem
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0
$$
Next, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Fina... | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,664 |
4. In quadrilateral $A B C D$, the diagonals intersect at point $O$. It is known that $S_{A B O}=S_{C D O}=\frac{3}{2}$, $B C=3 \sqrt{2}$, $\cos \angle A D C=\frac{3}{\sqrt{10}}$. Find the smallest area that such a quadrilateral can have. | Solution. We will prove that quadrilateral $ABCD$ is a parallelogram. Let $x_{1}, x_{2}, y_{1}, y_{2}$ be the segments into which the diagonals are divided by their point of intersection. Denote the angle between the diagonals as $\alpha$. By the condition, the areas of triangles $ABO$ and $CDO$ are equal, that is, $\f... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,665 |
1. Solve the equation $x^{2}-2 x \cdot \sin (x \cdot y)+1=0$. | Solution. We will interpret the left side of the equation as a quadratic trinomial in terms of $x$. For the roots to exist, the discriminant must be non-negative, i.e., $D=4 \sin ^{2}(x \cdot y)-4 \geq 0 \Leftrightarrow \sin ^{2}(x \cdot y)=1 \Leftrightarrow \cos 2 x y=-1 \Leftrightarrow x y=\frac{\pi}{2}+\pi n, n \in ... | (\1,\frac{\pi}{2}+2\pik),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,666 |
2. The natural numbers from 1 to 100 were written in a row without spaces. Then, plus signs were placed between some of the digits. (For example, $1234567+891011 \ldots 15+1617$... 99100.) Can the resulting sum be divisible by 111? | Solution. The remainder of a number divided by three is equal to the remainder of the sum of its digits divided by three. Therefore, the sum of the remainders of all numbers divided by three, separated by plus signs, is equal to the remainder of the sum of all digits of numbers from 1 to 100 divided by 3. In turn, this... | no | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,667 |
3. Compare the numbers $\left(10^{2017}+10^{2016}+\cdots+10+1\right)^{2018}$ and $\quad\left(10^{2018}+10^{2017}+\cdots+10+\right.$ $1)^{2017}$. | Solution. Let
$$
A=\left(10^{2017}+10^{2016}+\cdots+10+1\right)^{2018}, \quad B=\left(10^{2018}+10^{2017}+\cdots+10+1\right)^{2017}
$$
Using the formula for the sum of terms of a geometric progression, we find:
$$
A=\frac{\left(10^{2018}-1\right)^{2018}}{9^{2018}}, \quad B=\frac{\left(10^{2019}-1\right)^{2017}}{9^{2... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,668 |
4. The base of the triangular pyramid $S A B C$ is an equilateral triangle $A B C$ with a side length of 4. It is known that for any point $M$ on the extension of the height of the pyramid $S H$ (point $S$ is between points $M$ and $H$), the angles $M S A, M S B, M S C, A S B, A S C$, and $B S C$ are equal to each othe... | Solution. Let's choose point $M$ (from the problem statement) such that $M A=4$. Then point $S$ will be the center of the circumscribed sphere around the regular tetrahedron $M A B C$. Consequently, each of the four identical pyramids $S A B C, S M A B, S M A C, S A B C$ will cut off an equal part from the sphere of ra... | \pi\cdot\frac{7\sqrt{6}-9}{27} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,669 |
5. Prove that for any natural number $n$ there exists a natural number $N$ such that the product $9 \cdot 5^{n} \cdot N$ is a palindrome, that is, a number whose decimal representation reads the same from right to left as from left to right. For example, for $n=1$ we can take $N=13$, since $9 \cdot 5^{1} \cdot 13=585$. | Solution. We will use the following fact. Let a natural number $B$ be divisible by $5^{n}$, and its decimal representation contains at least $n$ digits, that is, $B=b_{m-1} b_{m-2} \cdots b_{1} b_{0}$ and $m \geq n$. Then, by adding arbitrary digits to the left of the decimal representation of the number $B$, we will a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,670 |
6. $\quad$ For what values of the parameter $a$ does the system of equations
$$
\left\{\begin{array}{c}
\sqrt{(x-6)^{2}+(y-13)^{2}}+\sqrt{(x-18)^{2}+(y-4)^{2}}=15 \\
(x-2 a)^{2}+(y-4 a)^{2}=\frac{1}{4}
\end{array}\right.
$$
have a unique solution
# | # Solution.
The first equation of the system defines the locus of points $M(x, y)$ on the plane, the sum of whose distances from points $A(6,13)$ and $B(18,4)$ is 15. Note that
$$
|A B|=\sqrt{12^{2}+9^{2}}=15
$$
Therefore, according to the triangle inequality, the locus of such points $M(x, y)$ is the segment $A B$.... | =\frac{145}{44},=\frac{135}{44},\in(\frac{63}{20},\frac{13}{4}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,671 |
7. A circle inscribed in a trapezoid intersects its diagonals at points $A, B, C, D$. Prove that the sum of the lengths of arcs $\overline{B A}+\overline{D C}$ is greater than the sum of the lengths of arcs $\overline{A D}+\overline{C B}$. | Solution. Let $O$ be the point of intersection of the diagonals.
It is known that the measure of angle $A O D$ is half the sum of the angular measures of arcs $\overline{C B}$ and $\overline... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,672 |
8. It is known that for any natural number $n$ the following formula holds:
$$
\cos (n \alpha)=2^{n-1} \cdot(\cos \alpha)^{n}+a_{n-1} \cdot(\cos \alpha)^{n-1}+a_{n-2} \cdot(\cos \alpha)^{n-2}+\cdots+a_{1} \cdot(\cos \alpha)+a_{0}
$$
Here $a_{k}$ are integers, and $a_{0}=0$ for odd $n$. Prove that for $n \geq 4$, the ... | Solution. In the proof, we will use the following statement: if a rational number $t=\frac{p}{q}$ (where $p$ and $q$ are coprime integers) is a root of the polynomial $f(x) = a_{k} x^{k} + a_{k-1} x^{k-1} + \cdots + a_{1} x + a_{0}$ with integer coefficients ($a_{i} \in \mathbb{Z}$), then $p$ is a divisor of $a_{0}$, a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,673 |
2. There are 5 cells. Two players take turns writing one of the digits 1 or 2 in these cells from left to right. If the resulting $5$-digit number is divisible by 3, the player who made the first move wins; if it is not divisible by 3, the second player wins. Which player will win with correct play and any play by the ... | # Solution:
Let's recall the divisibility rule for 3: an integer is divisible by 3 if and only if the sum of all its digits is divisible by 3.
We will outline a winning strategy for the second player. On each move, if the first player writes 1, the second player writes 2. If the first player writes 2, the second play... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,674 |
4. The bases of trapezoid $A B C D$ are related by the ratio $A D=4 \cdot B C$, and the sum of angles $\angle A+\angle D=120^{0}$. Points $M$ and $N$ are chosen on the lateral sides such that $C N: N D=B M: M A=1: 2 . \quad$ Perpendiculars drawn from points $M$ and $N$ to the lateral sides of the trapezoid intersect at... | # Solution:
Extend the lateral sides until they intersect at point $S$. The angle $S$
is $60^{\circ}$. Triangles $A S D$ and $B S C$ are similar; the similarity ratio is 4. Moreover, by the... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,675 |
6. Let $a_{n, m}$ denote the number obtained by writing down all numbers from $n$ to $m$ consecutively, where $n$ and $m$ are natural numbers, and $n > m \geq 1$. For example, the number $a_{4,2}=432$, and the number $a_{11,7}=1110987$. Prove that among such numbers, there is a number divisible by 2021. | # Solution:
Consider numbers of the form $a_{n, 1}$. Since there are infinitely many such numbers, there will be two numbers $a_{n, 1}$ and $a_{k, 1}, n>k$, that have the same remainder when divided by 2021. Then the difference $a_{n, 1}-a_{k, 1}$ is divisible by 2021. At the same time, $a_{n, 1}-a_{k, 1}=a_{n, k+1} \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,676 |
7. Prove that the system of equations has no solutions:
$$
\left\{\begin{array}{c}
x^{3}+x+y+1=0 \\
y x^{2}+x+y=0 \\
y^{2}+y-x^{2}+1=0
\end{array}\right.
$$ | # Solution:
Notice that $y\left(x^{3}+x+y+1\right)-x\left(y x^{2}+x+y\right)=y^{2}+y-x^{2}=0$, which contradicts the third equation of the system. | proof | Algebra | proof | Yes | Yes | olympiads | false | 8,677 |
8. Let's fix 10 natural numbers $n_{1}, n_{2}, \ldots, n_{10}$ and denote their sum by $n=n_{1}+\cdots+n_{10}$. Suppose now that on a board, $n$ numbers $a_{1}, \ldots, a_{n}$ are written in a row, each of which is either 0 or 1. These numbers (in the order they are written) are divided into 10 groups:
$$
\underbrace{... | # Solution:
The desired number of sets is found by summing the number of sets with a given number of non-zero groups $\mathrm{k}$:
For $\mathrm{k}=0$, such a set is unique;
For k=2 $\sum_{1 \leq i<j \leq 10}\left(2^{n_{i}}-2\right) \cdot\left(2^{n_{j}}-2\right)$;
for $\mathrm{k}=4 \sum_{1 \leq i<j<l<s \leq 10}\left(... | 2^{n-1}+\frac{1}{2}\cdot(2^{n_{1}}-2)\cdot(2^{n_{2}}-2)\cdot\ldots\cdot(2^{n_{10}}-2) | Combinatorics | proof | Yes | Yes | olympiads | false | 8,678 |
3. (10 points) On an 8x8 board, 9 checkers are placed. Movements of the checkers (moves) are carried out as follows: a first (movable) and a second checker are chosen. Then the first is placed on such a cell that the initial and final positions are symmetric relative to the second checker. Can such moves be used to mov... | Justify the answer
 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,679 | |
8. In a circle, three chords $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. The angular measures of the arcs $A C_{1}, A B, C A_{1}$, and $A_{1} B_{1}$ are $150^{\circ}, 30^{\circ}, 60^{\circ}$, and $30^{\circ}$, respectively. Find the angular measure of the arc $B_{1} C_{1}$. | Solution: Let's formulate several auxiliary statements.
1) Let the angular measure of the arc $AB$ (Fig.1) be $\varphi$. (This means that $\varphi$ is equal to the corresponding central angle $AOB$.) Then the length of the chord
$ and $2 \cdot\left(180^{\circ}-\angle B\right)$. The sum of the degree measures of arcs $\overline{A D}, \widetilde{C A}$, and $\overline{D C}$ is $360^{\circ}$. Therefore, the measu... | CD=R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,683 |
2. The function $y=f(x)$ is defined on the set $(0,+\infty)$ and takes positive values on it. It is known that for any points $A$ and $B$ on the graph of the function, the areas of the triangle $A O B$ and the trapezoid $A B H_{B} H_{A}$ are equal to each other ( $H_{A}, H_{B}$ - the bases of the perpendiculars dropped... | # Solution:
Let $M$ be the intersection point of segments $O B$ and $A H_{A}$. Since the areas of triangle $A O B$ and trapezoid $A B H_{B} H_{A}$ are equal, the areas of triangles $A M O$ and trapezoid $M B H_{B} H_{A}$ are also equal.
=1-x-4 x^{2}+x^{4}$ and $g(x)=16-8 x-$ $16 x^{2}+x^{4}$ respectively. Find $\frac{x_{2}}{x_{1}}$. | # Solution:
Notice that $f(-2)>0, f(-1)<0, f(0)>0, f(1)<0$. Therefore, the polynomial $f(x)$ has 4 real roots. Similarly, from the inequalities $g(-4)>0, g(-2)<0, g(0)>0, g(2)<0$, it follows that the polynomial $g(x)$ has 4 real roots.
Comparison of the coefficients of the polynomials
$$
f(x)=1-x-4 x^{2}+x^{4} \text... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,685 |
4. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$
# | # Solution:
Consider the strictly increasing sequence of values:
$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$
If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness ... | 43 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,686 |
5. It is known that the number $\cos 6^{0}$ is a root of the equation $32 t^{5}-40 t^{3}+10 t-\sqrt{3}=0$. Find the other four roots of this equation. (Answers in the problem should be compact expressions, not containing summation signs, ellipses, and radicals.)
# | # Solution:
$\cos 5 x=16 \cos ^{5} x-20 \cos ^{3} x+5 \cos x$. Then the other roots of the equation will be the numbers $\cos 66^{\circ}, \cos 78^{\circ}, \cos 138^{\circ}, \cos 150^{\circ}$.
Answer: $\cos 66^{\circ}, \cos 78^{\circ}, \cos 138^{\circ}, \cos 150^{\circ}$ | \cos66,\cos78,\cos138,\cos150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,687 |
1. (3 points) Oleg usually arrives on a business trip by the 11 AM train. A car arrives at the station for this train. This time, the train arrived an hour earlier, and Oleg started walking towards the car. Meeting the car on the way, he got in, and as a result, arrived 10 minutes earlier than planned. Determine the ti... | 1. 10 hours 55 minutes.
+ (3 points) - the solution is correct (by any method)
$+-(2$ points) - the solution is correct, but there are arithmetic errors
-+ (1 point) - there are reasonable ideas in solving the problem, but the problem is not solved overall | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,688 |
2. (3 points) Find all solutions of the inequality $\cos 5+2 x+x^{2}<0$, lying in the interval $\left[-2 ;-\frac{37}{125}\right]$. | 2. $x \in\left(-1-\sqrt{1-\cos 5} ;-\frac{37}{125}\right]$
+ (3 points) - solution is correct
+- (2 points) - solution is correct, but in the process of comparing $-1+\sqrt{1-\cos 5}$ with $-\frac{37}{125}$
not all steps are sufficiently justified (for example, it is not explained why $\left.\cos 5<\frac{37}{125} \c... | x\in(-1-\sqrt{1-\cos5};-\frac{37}{125}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,689 |
3. (4 points) In a parallelogram with sides 4 and 7, the bisectors of the four interior angles are drawn. Find the ratio of the area of the quadrilateral formed by the intersection of the bisectors to the area of the parallelogram. | 3. $\frac{S_{Q R T F}}{S_{A B C D}}=\frac{9}{56}$.
+ (4 points) - solution is correct.
+- (3 points) - solution is correct, but there are minor inaccuracies.
-+ (2 points) - solution is correct, but there are significant gaps (for example, it is not justified that the angle bisectors form a rectangle). | \frac{9}{56} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,690 |
4. (4 points) Determine all values of the parameter $a$ for which the equation $x+\sqrt{a-x^{2}}=1$ has two distinct solutions. | 4. For $a \in\left(\frac{1}{2} ; 1\right]$ the equation has two distinct solutions.
$+(4$ points) - the solution is correct.
+- (3 points) - the solution is generally correct, but there are minor inaccuracies in selecting the parameter values.
-+ (2 points) - the roots $x=\frac{1 \pm \sqrt{2 a-1}}{2}$ were found, bu... | \in(\frac{1}{2};1] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,691 |
5. (5 points) It is known that natural numbers $a, b$ satisfy two conditions:
- the sum of $a$ and $b$ is 546,
- the least common multiple of $a$ and $b$ is 22 times their greatest common divisor.
Find $a$ and $b$. | 5. 462 and 84, or 84 and 462.
$+(5$ points) - the solution is correct.
+- (3 points) - the solution is generally correct, but not all cases for numbers $a$ and $b$ are considered.
-+ (2 points) - in the solution, the correct conclusion is drawn from the relation $\frac{[a, b]}{(a, b)}=22$ about the canonical decompo... | 46284,or84462 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,692 |
1. Find the sum of all even natural numbers $n$ for which the number of divisors (including 1 and $n$ itself) is equal to $\frac{n}{2}$. (For example, the number 12 has 6 divisors: $1,2,3,4,6,12$.) | Solution. Let the canonical decomposition of the number $n$ be: $n=2^{t_{1}} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdots \cdot p^{t_{k}}$. Then the number of divisors of the number $n$ is $\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)$. From the condition of the problem, we h... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,693 |
2. In how many ways can 4 numbers be chosen from the first 1000 natural numbers $1,2, \ldots, 1000$ to form an increasing arithmetic progression? | Solution. Let's find the formula for calculating the number of ways to choose 4 numbers from the first $n$ natural numbers $1,2, \ldots, n$ that form an increasing arithmetic progression. The number of progressions with a difference of 1 is $n-3$ (the first term of the progression can take values from 1 to $n-3$ inclus... | 166167 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,694 |
3. A motorist set off from point A to point D, the distance between which is 100 km. The road from A to D passes through points B and C. At point B, the navigator showed that 30 minutes of driving remained, and the motorist immediately reduced the speed by 10 km/h. At point C, the navigator showed that 20 km remained t... | Solution. According to the condition, the distance from C to D is 20 km. Let the distance from A to B be $x$ (km), then the distance from B to C will be ( $80-x$ ) km.
Let $v \frac{\text { km }}{4}$ be the initial speed of the car. Then on the segments BC and CD, it is ( $v-$
^{N}=4817152-2781184 \cdot \sqrt{3}$. Find $N$. | Solution. Suppose that raising the number $a+b \sqrt{3}$ to the power $N$ results in the number $A+B \sqrt{3}$ (where $a, b, A, B$ are integers). Expanding the expression $(a+b \sqrt{3})^{N}$, we get a sum of monomials (with non-essential (for us now) integer coefficients) of the form $a^{N-n}(b \sqrt{3})^{n}$. The ter... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,696 |
5. In triangle $\mathrm{ABC}$, the sides $A B=4, B C=6$. Point $M$ lies on the perpendicular bisector of segment $A B$, and lines $A M$ and $A C$ are perpendicular. Find $M A$, if the radius of the circumscribed circle around triangle $A B C$ is 9. | Solution. Introduce a coordinate system with the origin at point A such that point C lies on the x-axis. From the problem statement, point M lies on the y-axis. Let's introduce notations for the unknown coordinates: $\mathrm{A}(0,0), \mathrm{B}\left(\mathrm{x}_{\mathrm{B}}, \mathrm{y}_{\mathrm{B}}\right), \mathrm{C}\le... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,697 |
1. Find all functions $f(x)$ that simultaneously satisfy the following three conditions:
1) $f(x)>0$ for any $x>0$;
2) $f(1)=1$;
3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b) + a^{2} + b^{2}$ for any $a, b \in \mathbb{R}$. | Solution. In the identity from the condition
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0
$$
Then, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Finall... | f(x)=x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,698 |
3. Anya and Borya are playing "battleship" according to the following rules: 29 different points are chosen on a circle, numbered clockwise with natural numbers from 1 to 29. Anya draws a ship - an arbitrary triangle with vertices at these points. Borya (not knowing the location of Anya's ship) makes a "shot": he names... | Solution: There are a total of $C_{29}^{3}=3654$ different triangles. One shot "wounds" 27 triangles. By making 134 shots, no more than $134 \cdot 27=3618$ triangles can be "wounded". Since $C_{29}^{3}>134 \cdot 27$, 134 shots are not enough to guarantee that the ship will be "wounded".[^0]
## Interregional Olympiad f... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,700 |
4. It is known that the equation $x^{3}-x-1=0$ has a unique real root $x_{0}$. Come up with at least one equation of the form
$$
a \cdot z^{3}+b \cdot z^{2}+c \cdot z+d=0
$$
where $a, b, c, d$ are integers and $a \neq 0$, one of the roots of which is the number
$$
z=x_{0}^{2}+3 \cdot x_{0}+1
$$ | Solution: Let's write down the relations
$$
\begin{gathered}
z=x_{0}^{2}+3 \cdot x_{0}+1 \\
z \cdot x_{0}=x_{0}^{3}+3 \cdot x_{0}^{2}+x_{0} \\
z \cdot x_{0}^{2}=x_{0}^{4}+3 \cdot x_{0}^{3}+x_{0}^{2}
\end{gathered}
$$
The right-hand sides can be simplified (reduced modulo $x_{0}^{3}-x_{0}-1$), using the fact that $x_{... | z^{3}-5z^{2}-10z-11=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,701 |
5. In quadrilateral $A B C D$, the diagonals intersect at point $O$. It is known that
$S_{A B O}=S_{C D O}=\frac{3}{2}, B C=3 \sqrt{2}, \cos \angle A D C=\frac{3}{\sqrt{10}}$. Find the sine of the angle between the diagonals of this quadrilateral, if its area takes the smallest possible value under the given condition... | Solution. We will prove that quadrilateral $ABCD$ is a parallelogram. Let $x_{1}, x_{2}, y_{1}, y_{2}$ be the segments into which the diagonals are divided by their point of intersection. Denote the angle between the diagonals as $\alpha$. By the condition, the areas of triangles $ABO$ and $CDO$ are equal, that is, $\f... | \frac{6}{\sqrt{37}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,702 |
6. Find all prime numbers whose representation in base 14 has the form 101010 ... 101 (ones and zeros alternate). | Solution: Let $2n+1$ be the number of digits in the number $A=101010 \ldots 101$. Let $q=$ 14 - the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2 n}=\frac{q^{2 n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2 k \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k+1}-1}{q-1} \cdot \frac... | 197 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,703 |
7. Prove that for all $x \in\left(0, \frac{3 \pi}{8}\right)$ the inequality holds:
$$
\frac{1}{\sin \frac{x}{3}}+\frac{1}{\sin \frac{8 x}{3}}>\frac{\sin \frac{3 x}{2}}{\sin \frac{x}{2} \sin 2 x}
$$
Hint: Use the concavity of the graph of the function $f(t)=\frac{1}{\sin t}$ on the interval $(0 ; \pi)$ | Solution. Let's perform the transformations
$$
\begin{aligned}
& \frac{1}{\sin \frac{x}{3}}+\frac{1}{\sin \frac{8 x}{3}}>\frac{2 \cos \frac{x}{2} \sin \frac{3 x}{2}}{2 \cos \frac{x}{2} \sin \frac{x}{2} \sin 2 x} \Leftrightarrow \frac{1}{\sin \frac{x}{3}}+\frac{1}{\sin \frac{8 x}{3}}>\frac{2 \cos \frac{x}{2} \sin \frac... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,704 |
8. In each of the $k$ cells of an $n \times n$ square table, a one is written, and in the remaining cells - a zero. Find the maximum value of $k$, such that regardless of the initial placement of the ones, by swapping rows with each other and columns with each other, it is possible to achieve that all ones are above th... | Solution: We will denote a table of size $n \times n$ as $\mathbf{T}_{n}$. It is obvious that for the table $\mathbf{T}_{2}$, the maximum $k$ is 3. Experimenting with the table $\mathbf{T}_{3}$, we can notice that $k=4$ (we will prove this rigorously below). These observations allow us to hypothesize that for any $n>1$... | n+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,705 |
1. How many solutions of the equation $x^{2}-2 x \cdot \sin (x \cdot y)+1=0$ fall within the circle $x^{2}+y^{2} \leq 100$? | Solution. We will interpret the left side of the equation as a quadratic trinomial in terms of $x$. For the roots to exist, the discriminant must be non-negative, i.e., $D=4 \sin ^{2}(x \cdot y)-4 \geq 0 \Leftrightarrow \sin ^{2}(x \cdot y)=1 \Leftrightarrow \cos 2 x y=-1 \Leftrightarrow x y=\frac{\pi}{2}+\pi n, n \in ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,706 |
3. It is known that the polynomial $f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coefficient $b_{1}$... | Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$
$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}
$$
Then, according to the problem, we have
$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x... | -1216 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,707 |
4. Find the smallest value of the parameter $a$, for which the system of equations
$$
\left\{\begin{array}{c}
\sqrt{(x-6)^{2}+(y-13)^{2}}+\sqrt{(x-18)^{2}+(y-4)^{2}}=15 \\
(x-2 a)^{2}+(y-4 a)^{2}=\frac{1}{4}
\end{array}\right.
$$
has a unique solution
# | # Solution.
The first equation of the system defines the locus of points $M(x, y)$ on the plane for which the sum of the distances from these points to points $A(6,13)$ and $B(18,4)$ is 15. Note that
$$
|A B|=\sqrt{12^{2}+9^{2}}=15
$$
Therefore, according to the triangle inequality, the locus of such points $M(x, y)... | \frac{135}{44} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,708 |
1. A robot is located in one of the cells of an infinite grid and can be given the following commands:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the ... | Solution. For brevity, let's denote the command to the left as L, to the right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,709 |
2. We have a pencil, a ruler, and a certain special device that, for any angle drawn on a plane, constructs two rays that divide this angle into three equal angles. Using these tools, construct an angle of $10^{0}$ on the plane. (Let's recall that a pencil can mark a point on the plane, in particular, the intersection ... | Solution. On the line passing through two distinct points $A$ and $D$, mark points $B$ and $C$ as shown in the figure. Divide the straight angles $ABC$ and $DCB$ (which should be imagined as initially laid out in the upper, and then in the lower half-plane relative to the line $AD$) into three equal parts. As a result,... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,710 |
3. Real numbers $x, y, z$ satisfy the relations:
$$
4 x^{2}-2 x-30 y z=25 y^{2}+5 y+12 x z=9 z^{2}-3 z-20 x y .
$$
Find all possible triples of numbers $(a, b, c)$, where $a=2 x+5 y, b=3 z+5 y, c=3 z-2 x$. | Solution. Note that
$$
a-b+c=0
$$
## Interregional School Olympiad in Mathematics Based on Departmental Educational Organizations
Let $A=4 x^{2}-2 x-30 y z, B=25 y^{2}+5 y+12 x z$ and $C=9 z^{2}-3 z-20 x y$. Subtracting these equations from each other, we get:
$$
\begin{aligned}
& A-B=a \cdot(2 x-6 z-5 y-1)=0 \\
& ... | (0,0,0),(0,1,1),(1,0,-1),(-1,-1,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,711 |
4. Find all functions $f(x)$ that simultaneously satisfy the following three conditions: 1) $f(x)>0$ for any $x>0$; 2) $f(1)=1$; 3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}$ for any $a, b \in \mathbb{R}$. | Solution. In the identity from the condition
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0.
$$
Then, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Final... | f(x)=x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,712 |
5. In quadrilateral $A B C D$, the diagonals intersect at point $O$. It is known that $S_{A B O}=S_{C D O}=\frac{3}{2}$, $B C=3 \sqrt{2}$, and $\cos \angle A D C=\frac{3}{\sqrt{10}}$. Find the sine of the angle between the diagonals of this quadrilateral, if its area takes the smallest possible value under the given co... | Solution. We will prove that quadrilateral $ABCD$ is a parallelogram. Let $x_{1}, x_{2}, y_{1}, y_{2}$ be the segments into which the diagonals are divided by their point of intersection. Denote the angle between the diagonals as $\alpha$. By the given condition, the areas of triangles $ABO$ and $CDO$ are equal, that i... | \frac{6}{\sqrt{37}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,713 |
6. Find all prime numbers whose decimal representation has the form 101010 ... 01. | Solution. Let $2 n+1$ be the number of digits in the number $A=101010 \ldots 101$ under investigation. Let $q=$ 10 be the base of the number system. Then $A=q^{0}+q^{2}+\cdots q^{2 n}=\frac{q^{2 n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2 k \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k+1... | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,714 |
7. The ordinary fraction $\frac{1}{221}$ is represented as a periodic decimal fraction. Find the length of the period. (For example, the length of the period of the fraction $\frac{25687}{99900}=0.25712712712 \ldots=0.25$ (712) is 3.) | Solution. Let's consider an example. We will convert the common fraction $\frac{34}{275}$ to a decimal. For this, we will perform long division (fig.). As a result, we find $\frac{34}{275}=0.123636363 \ldots=0.123(63)$. We obtain the non-repeating part 123 and the repeating part 63. Let's discuss why the non-repeating ... | 48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,715 |
8. Anya and Borya are playing "battleship" with the following rules: 29 different points are chosen on a circle, numbered clockwise with natural numbers from 1 to 29. Anya draws a ship - an arbitrary triangle with vertices at these points. We will call a "shot" the selection of two different natural numbers $k$ and $m$... | Solution. The vertices of Anya's triangle divide the circle into three arcs (see figure). Let $x, y$, and $26-x-y$ be the number of points on these arcs (see figure), excluding the vertices of the triangle itself. For a "shot" with endpoints at points $k$ and $m$ to not hit the ship, both these points must lie on one o... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,716 |
1. (3 points) Given 10 different natural numbers, each not exceeding 23. Prove that there exist four different numbers $a, b, c, d$ among them, such that $\frac{a+b}{2}=\frac{c+d}{2}$. | # Solution:
Let $a_{1}, \ldots, a_{10}$ be natural numbers $\leq 23$. There are $9+8+\ldots+2+1=\frac{9 \cdot 10}{2}=45$ different pairs of numbers $\left(a_{1}, a_{2}\right),\left(a_{1}, a_{3}\right), \ldots,\left(a_{9}, a_{10}\right)$. For each pair, the sum of the numbers in it is greater than 2 and less than 46. T... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,717 |
2. (3 points) The machine works with a magnetic card, on which any pair of natural numbers can be recorded. With the record $(m ; n)$, it can perform any of the following actions:
4) swap the numbers: $(m ; n) \rightarrow(n ; m)$;
5) replace the first number with the sum of the first and second: $(m ; n) \rightarrow(m+... | # Solution:
It can be noticed that all operations performed by the automaton preserve the GCD of the pair of numbers presented to it.
The numbers given in the condition can be factored into prime factors as follows: $\quad 1037=17 \times 61, \quad 1159=19 \times 61, \quad 611=13 \times 47$, $1081=23 \times 47$.
Thus... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,718 |
3. (4 points) Solve the equation of the form $f(f(x))=x$, given that $f(x)=x^{2}+5 x+1$.
# | # Solution:
It is clear that some solutions to the equation $f(f(x))=x$ can be found by solving the simpler equation $f(x)=x$, which in this case is:
$$
x^{2}+5 x+1=x \Leftrightarrow x^{2}+4 x+1=0 \Leftrightarrow x=-2 \pm \sqrt{3} \text {. }
$$
Since the roots of the polynomial $f(x)-x=x^{2}+4 x+1$ are also roots of... | -2\\sqrt{3},-3\\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,719 |
4. (4 points) Numbers are marked on a chessboard (see Fig. 1). How many arrangements of 8 rooks, none of which attack each other, exist such that the numbers on the squares occupied by the rooks include all numbers from 0 to 7? $\left(\begin{array}{llllllll}0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7... | # Solution:
Let's number the rows of the board from 1 to 8 from top to bottom. On the first row, the position of the rook can be chosen in 8 ways. For example, we choose the first cell with the number 0. On the second row, there are 7 options left. Six of them (cells from the second to the seventh) have the property t... | 3456 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,720 |
5. (5 points) Solve the system of equations:
$$
\left\{\begin{array}{c}
x^{3}+4 y=y^{3}+16 x \\
\frac{y^{2}+1}{x^{2}+1}=5
\end{array}\right.
$$ | # Solution:
The following equivalent transformations are valid:
$$
\left\{\begin{array} { c }
{ x ^ { 3 } + 4 y = y ^ { 3 } + 1 6 x } \\
{ \frac { y ^ { 2 } + 1 } { x ^ { 2 } + 1 } = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
y^{2}=5 x^{2}+4 \\
x^{3}+4 y=y^{3}+16 x
\end{array} \Leftrightarrow\right.\rig... | (0;\2),(\1;\3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,721 |
6. (5 points) From point $A$, lying on a circle of radius 3, chords $A B, A C$ and tangent $A D$ are drawn. The angle between the chords is $\frac{\pi}{4}$, and the angle between chord $A C$ and tangent $A D$, which does not contain chord $A B$, is $\frac{5 \pi}{12}$. Calculate the integer area of triangle $A B C$.
# | # Solution:
Let $\angle D A C=\alpha, \angle B A C=\beta$, and the radius of the circle be $R$. It is known that $\angle A C B=\angle D A C=\alpha$.
By the Law of Sines, $\quad \frac{|A B|}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,722 |
1. (3 points) Given 11 different natural numbers, each not exceeding 27. Prove that there exist four different numbers $a, b, c, d$ among them, such that $\frac{a+b}{5}=\frac{c+d}{5}$. | # Solution:
Let $a_{1}, \ldots, a_{11}$ be natural numbers $\leq 27$. We have
$10+9+8+\ldots+2+1=\frac{11 \cdot 10}{2}=55$ different
numbers
$\left(a_{1}, a_{2}\right),\left(a_{1}, a_{3}\right), \ldots,\left(a_{10}, a_{11}\right)$. For each pair, the sum of the numbers in it is greater than 2 and less than 54. Ther... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,723 |
2. (3 points) The machine works with a magnetic card, on which any pair of natural numbers can be recorded. With the recording ( $m ; n$ ), it can perform any of the following actions:
4) swap the numbers: $(m ; n) \rightarrow(n ; m)$;
5) replace the first number with the sum of the first and second: $(m ; n) \rightarr... | # Solution:
It can be observed that all operations performed by the machine preserve the GCD of the pair of numbers presented to it.
The numbers given in the condition can be factored into prime factors as follows: $901=17 \times 53,1219=23 \times 53,871=13 \times 67,1273=19 \times 67$.
Thus, the GCD of the first pa... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,724 |
3. (4 points) Solve the equation of the form $f(f(x))=x$, given that $f(x)=x^{2}+2 x-5$
# | # Solution:
It is clear that some solutions to the equation $f(f(x))=x$ can be found by solving the simpler equation $f(x)=x$, which in this case is:
$$
x^{2}+2 x-5=x \Leftrightarrow x^{2}+x-5=0 \Leftrightarrow x=\frac{1}{2}(-1 \pm \sqrt{21})
$$
Since the roots of the polynomial $f(x)-x=x^{2}+x-5$ are also roots of ... | \frac{1}{2}(-1\\sqrt{21}),\frac{1}{2}(-3\\sqrt{17}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,725 |
5. (5 points) Solve the system of equations:
$$
\left\{\begin{array}{c}
x^{3}+2 y=y^{3}+14 x \\
\frac{y^{2}+1}{x^{2}+1}=3
\end{array}\right.
$$ | # Solution:
The following equivalent transformations are valid:
$$
\left\{\begin{array} { c }
{ x ^ { 3 } + 2 y = y ^ { 3 } + 1 4 x } \\
{ \frac { y ^ { 2 } + 1 } { x ^ { 2 } + 1 } = 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
y^{2}=3 x^{2}+2 \\
x^{3}+2 y=y^{3}+14 x
\end{array} \Leftrightarrow\right.\rig... | (0;\\sqrt{2}),(\\sqrt{2};\2\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,727 |
6. (5 points) From point $A$, lying on a circle, chords $A B$, $A C$, and tangent $A D$ are drawn. The angle between the chords is $\frac{\pi}{6}$, and the angle between chord $A C$ and tangent $A D$, which does not contain chord $A B$, is $\frac{5 \pi}{12}$. Calculate the integer part of the radius of the circle if th... | # Solution:
Let $\angle D A C=\alpha, \angle B A C=\beta$, and the radius of the circle be $R$. It is known that $\angle A C B=\angle D A C=\alpha$.
By the Law of Sines, $\quad \frac{|A B|}... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,728 |
8. There is an unlimited number of test tubes of three types - A, B, and C. Each test tube contains one gram of a solution of the same substance. Test tubes of type A contain a $10\%$ solution of this substance, type B $-20\%$ solution, and type C $-90\%$ solution. Sequentially, one after another, the contents of the t... | Solution: Let the number of test tubes of types A, B, and C be $a$, $b$, and $c$ respectively. According to the problem, $0.1a + 0.2b + 0.9c = 0.2017 \cdot (a + b + c) \Leftrightarrow 1000 \cdot (a + 2b + 9c) = 2017 \cdot (a + b + c)$. The left side of the last equation is divisible by 1000, so the right side must also... | 73 | Other | math-word-problem | Yes | Yes | olympiads | false | 8,730 |
9. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $1^{2}+2^{2}+4^{2}=21$ ). | Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Co... | 5035485 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,731 |
11. In a triangle with sides $a, b, c$ and angles $\alpha, \beta, \gamma$, the equality $3 \alpha + 2 \beta = 180^{0}$ is satisfied. The sides $a, b, c$ lie opposite the angles $\alpha, \beta, \gamma$ respectively. Find the length of side $c$ when $a=2, b=3$.
 the matrix has the form $\left(\begin{array}{lll}1 & * & * \\ * & 1 & * \\ * & * & 1\end{array}\right)$, where each * can take the value 0 or 1
2) the rows of the matrix do not repeat. | Solution: Let $A$ be the set of matrices satisfying condition 1) and $B$ be the subset of $A$ consisting of matrices satisfying condition 2). We need to find the number of elements in the set $B$. Let $A_{ij}, i, j \in\{1,2,3\}, i \neq j$, be the subset of $A$ consisting of matrices in which row $i$ and $j$ are the sam... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,734 |
1. (2 points) For a natural number $x$, six statements are made:
$$
\begin{aligned}
& 3 x>91 \\
& x<37 \\
& 2 x \geq 21 \\
& x>7
\end{aligned}
$$
It is known that only three of them are true, and three are false. Find $x$. | Solution. Let's transform the original system of inequalities
$$
\begin{aligned}
& x>\frac{91}{3} \\
& x \geq \frac{21}{2} \\
& x>\frac{37}{4} \\
& x>7 \\
& x<27 \\
& x<120
\end{aligned}
$$
The first inequality does not hold, because otherwise four inequalities would be satisfied immediately. Therefore, $x \leq \frac... | x\in{8,9} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,736 |
3. (4 points) Solve the equation $\frac{\sin ^{2} 3 x}{\sin ^{2} x}=8 \cos 4 x+\frac{\cos ^{2} 3 x}{\cos ^{2} x}$. | Solution: O.D.Z. $x \neq \frac{\pi}{2} k, k \in Z$.
$$
\begin{aligned}
& \frac{\sin ^{2} 3 x}{\sin ^{2} x}=8 \cos 4 x+\frac{\cos ^{2} 3 x}{\cos ^{2} x} \Leftrightarrow \\
& \frac{\sin ^{2} 3 x \cos ^{2} x-\sin ^{2} x \cos ^{2} 3 x}{\sin ^{2} x \cos ^{2} x}=8 \cos 4 x \Leftrightarrow \frac{\sin 2 x \sin 4 x}{\frac{1}{4... | \\frac{\pi}{3}+\pi,\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,739 |
4. (4 points) When preparing for the exam, three schoolchildren solved 100 problems. Each schoolchild solved 60 problems, and every problem was solved by someone. A problem is considered difficult if only one schoolchild solved it. An easy problem is one that was solved by all three schoolchildren. Which type of proble... | Solution. Let
$x_{i}$ - the number of problems solved only by the $i$-th student,
$y_{i, j}$ - the number of problems solved only by the $i$-th and $j$-th students,
$z$ - the number of problems solved by all students (the number of easy problems)
Then the number of difficult problems is $x_{1}+x_{2}+x_{3}$.
By the... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,740 |
5. (5 points) Find all pairs of integers $(x, y)$ for which the following equality holds:
$$
x(x+1)(x+7)(x+8)=y^{2} .
$$ | Solution: $x(x+1)(x+7)(x+8)=y^{2} \quad \Leftrightarrow$
$$
\left(x^{2}+8 x\right)\left(x^{2}+8 x+7\right)=y^{2}
$$
Let $z=x^{2}+8 x+\frac{7}{2} \Rightarrow 2 z=2 x^{2}+16 x+7 \in Z$. From (2) we find
$$
\left(z-\frac{7}{2}\right)\left(z+\frac{7}{2}\right)=y^{2} \Leftrightarrow z^{2}-\frac{49}{4}=y^{2}
$$
which is ... | (1;12),(1;-12),(-9;12),(-9;-12),(0;0),(-8;0),(-4;-12),(-4;12),(-1;0),(-7;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,741 |
1. Compare the numbers
$$
\left(10^{2017}+10^{2016}+\cdots+10+1\right)^{2018} \text { and }\left(10^{2018}+10^{2017}+\cdots+10+1\right)^{2017} .
$$ | Solution. Let
$$
A=\left(10^{2017}+10^{2016}+\cdots+10+1\right)^{2018}, \quad B=\left(10^{2018}+10^{2017}+\cdots+10+1\right)^{2017}
$$
Using the formula for the sum of terms of a geometric progression, we find:
$$
A=\frac{\left(10^{2018}-1\right)^{2018}}{9^{2018}}, \quad B=\frac{\left(10^{2019}-1\right)^{2017}}{9^{2... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,743 |
3. In how many ways can 4 numbers forming an increasing arithmetic progression be chosen from the first $n$ natural numbers $1,2, \ldots, n$? | Solution. The number of progressions with a difference of 1 is $n-3$ (the first term of the progression can take values from 1 to $n-3$ inclusive), the number of progressions with a difference of 2 is $n-$ $6, \ldots$, the number of progressions with a difference of $d$ is $n-3 d$. The difference $d$ satisfies the ineq... | \frac{(2n-3k-3)k}{2},wherek=[\frac{n-1}{3}] | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,744 |
4. A motorist set off from point A to point B, the distance between which is 100 km. At the moment when the navigator showed that 30 minutes of travel remained, the motorist first reduced the speed by 10 km/h, and at the moment when the navigator showed that 20 km remained to travel, the motorist reduced the speed by t... | Solution. According to the condition, the distance from C to D is 20 km. Let the distance from A to B be denoted as
$x$ (km), then the distance from B to C will be $(80 - x)$ km. Let $v \frac... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,745 |
5. Solve the equation $\sin \frac{\pi n}{12} \cdot \sin \frac{\pi k}{12} \cdot \sin \frac{\pi m}{12}=\frac{1}{8}$. Here $k, m, n-$ are natural numbers not exceeding 5. | # Solution.
Let $1 \leq n \leq k \leq m \leq 5$.
Consider the cases:
1) $n=k=m=2$. Clearly, this set is a solution.
2) $\quad n=1$. Then note that for $k=2, m=5$ we get a valid equality.
The function $y=\sin x$ is increasing on the interval $\left(0 ; \frac{\pi}{2}\right)$, so the sets $\left.2 ; k ; m\right)$ for ... | (2;2;2),(1;2;5),(1;5;2),(2;1;5),(2;5;1),(5;1;2),(5;2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,746 |
6. In trapezoid $A B C D$ with bases $B C$ and $A D$, angle $D A B$ is a right angle. It is known that there exists a unique point $M$ on side $C D$ such that angle $B M A$ is a right angle. Prove that $B C=C M$ and $A D=M D$. | Solution. Construct a circle with side $AB$ as its diameter. Since angle $BMA$ is a right angle, point $M$ lies on this circle. Since such a point $M$ is unique on side $CD$, $CD$ is tangent to the circle. Therefore, $BC = CM$ and $AD = MD$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,747 |
7. The product of positive numbers $a$ and $b$ is greater than 1. Prove,
that for any natural number $n \geq 2$ the inequality
$$
(a+b)^{n}>a^{n}+b^{n}+2^{n}-2
$$
holds. | Solution. $(a+b)^{n}=a^{n}+t_{n-1} a^{n-1} b+t_{n-2} a^{n-2} b^{2}+\cdots t_{1} a b^{n-1}+b^{n}$, where $t_{k}$ are integers depending on $n$ and $k$, but not on $a$ and $b$, and at the same time
$t_{n-1}=t_{1}, t_{n-2}=t_{2}, \cdots, t_{n-k}=t_{k}, \cdots$
For $a=b=1$ we have: $2^{n}=1+t_{n-1}+t_{n-2}+\cdots t_{1}+1... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 8,748 |
1. Find all roots of the equation $\frac{1}{\cos ^{3} x}-\frac{1}{\sin ^{3} x}=4 \sqrt{2}$, lying in the interval $\left(-\frac{\pi}{2}, 0\right)$. Write the answer in degrees. | # Solution:
$\sin ^{3} x-\cos ^{3} x=4 \sqrt{2} \sin ^{3} x \cos ^{3} x \Leftrightarrow(\sin x-\cos x)\left(\sin ^{2} x+\sin x \cos x+\cos ^{2} x\right)=4 \sqrt{2} \sin ^{3} x \cos ^{3} x$. Substitution: $\sin x-\cos x=t, \sin x \cos x=\frac{1-t^{2}}{2}$. Then $t\left(3-t^{2}\right)=\sqrt{2}\left(1-t^{2}\right)^{3}$. ... | -45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,749 |
3. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $\left.1^{2}+2^{2}+4^{2}=21\right)$.
| Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Co... | 5035485 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,750 |
4. A circle touches the sides of an angle at points $A$ and $B$. A point $M$ is chosen on the circle. The distances from $M$ to the sides of the angle are 24 and 6. Find the distance from $M$ to the line $A B$.
=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coefficient $b_{1}$... | Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$: $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}$.
Then, according to the problem statement, we have:
$f(\quad)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\rig... | -1216 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,752 |
6. Find the number of matrices that satisfy two conditions:
3) the matrix has the form $\left(\begin{array}{lll}1 & * & * \\ * & 1 & * \\ * & * & 1\end{array}\right)$, where each * can take the value 0 or 1 and the rows of the matrix do not repeat. | Solution: Let $A$ be the set of matrices satisfying condition 1) and $B$ be the subset of $A$ consisting of matrices satisfying condition 2). We need to find the number of elements in the set $B$. Let $A_{ij}, i, j \in\{1,2,3\}, i \neq j$, be the subset of $A$ consisting of matrices in which row $i$ and $j$ are the sam... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,753 |
2. Non-zero numbers $a$ and $b$ are roots of the quadratic equation $x^{2}-5 p x+2 p^{3}=0$. The equation $x^{2}-a x+b=0$ has a unique root. Find $p$. | Solution. Since the equation $x^{2}-a x+b=0$ has a unique root, then $b=\frac{a^{2}}{4}$. By Vieta's theorem, we have the equalities: $a+b=5 p ; a b=2 p^{3}$. Substituting $b=\frac{a^{2}}{4}$ into the last equality, we get: $a=2 p$. Considering that $a$ and $b$ are non-zero, we find $p=3$.
Answer: 3. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,754 |
4. The function $y=f(x)$ is defined on the set $(0,+\infty)$ and takes positive values on this set. It is known that for any points $A$ and $B$ on the graph of the function, the areas of the triangle $A O B$ and the trapezoid $A B H_{B} H_{A}$ are equal to each other $\left(H_{A}, H_{B} - \text{the bases of the perpend... | Solution. Let $M$ be the intersection point of segments $O B$ and $A H_{A}$. Since the areas of triangle $A O B$ and trapezoid $A B H_{B} H_{A}$ are equal, the areas of triangles $A M O$ and trapezoid $M B H_{B} H_{A}$ are also equal. From this, it follows that the areas of triangles $A O H_{A}$ and trapezoid $B O H_{B... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,755 |
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