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6. A circle, the center of which lies on the line $y=b$, intersects the parabola $y=\frac{3}{4} x^{2}$ at least at three points; one of these points is the origin, and two of the remaining points lie on the line $y=\frac{3}{4} x+b$. Find all values of $b$ for which the described configuration is possible. | Answer: $b=\frac{25}{12}$.
Solution. First, consider $b>0$. Let the origin be denoted as $O(0 ; 0)$, the center of the circle as $Q(a ; b)$ (since it lies on the line $y=b$, its ordinate is $b$); the points of intersection of the line with the parabola as $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)\... | \frac{25}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,429 |
7. On the edges $A C, B C, B S, A S$ of a regular triangular pyramid $S A B C$ with vertex $S$, points $K, L, M, N$ are chosen respectively. It is known that points $K, L, M, N$ lie in the same plane, and $K L = M N = 2, K N = L M = 18$. In the quadrilateral $K L M N$, there are two circles $\Omega_{1}$ and $\Omega_{2}... | Answer: a) $\angle S A B=\arccos \frac{1}{6} ;$ b) $C Q=\frac{52}{3}$.
Solution. The opposite sides of quadrilateral $K L M N$ are pairwise equal, so it is a parallelogram. Since the plane $(K L M N)$ intersects the planes $(A B C)$ and $(A B S)$ along parallel lines $K L$ and $M N$, these lines are parallel to the li... | \angleSAB=\arccos\frac{1}{6};CQ=\frac{52}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,430 |
1. Given quadratic trinomials $f_{1}(x)=x^{2}-x-a, f_{2}(x)=x^{2}+b x+2, f_{3}(x)=4 x^{2}+(b-3) x-3 a+2$ and $f_{4}(x)=4 x^{2}+(3 b-1) x+6-a$. Let the differences of their roots be $A, B, C$ and $D$, respectively, and given that $|A| \neq|B|$. Find the ratio $\frac{C^{2}-D^{2}}{A^{2}-B^{2}}$. The values of $A, B, C, D,... | Answer: $\frac{1}{2}$.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=$ $\frac{\sqrt{T}}{|a|} \cdot... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,431 |
2. It is known that $\frac{\cos x-\sin x}{\sin y}=\frac{2 \sqrt{2}}{5} \tan \frac{x+y}{2}$ and $\frac{\sin x+\cos x}{\cos y}=-\frac{5}{\sqrt{2}} \cot \frac{x+y}{2}$. Find all possible values of the expression $\tan(x+y)$, given that there are at least three. | Answer: $-1, \frac{20}{21}$ or $-\frac{20}{21}$.
Solution. Multiplying the two given equations, we get $\frac{\cos ^{2} x-\sin ^{2} x}{\cos y \sin y}=-2$, which is equivalent to the following on the domain of definition:
$$
\cos 2 x=-2 \sin y \cos y \Leftrightarrow \cos 2 x+\cos \left(2 y-\frac{\pi}{2}\right)=0 \Left... | -1,\frac{20}{21},-\frac{20}{21} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,432 |
3. On the table, there are 140 different cards with numbers $4, 8, 12, \ldots, 556, 560$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the sum of the numbers on the selected cards is divisible by 3? | Answer: 149224.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 4. Therefore, the remainders of these numbers when divided by 3 alternate. Indeed, if one of these numbers is divisible by 3, i.e., has the form $3k$, where $k \in \mathbb{N}$, then the next nu... | 149224 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,433 |
4. Circles $\Omega$ and $\omega$ touch each other externally at point $F$, and their common external tangent touches circles $\Omega$ and $\omega$ at points $A$ and $B$, respectively. Line $\ell$ passes through point $B$, intersects circle $\omega$ again at point $C$, and intersects $\Omega$ at points $D$ and $E$ (poin... | Answer: $H P=\frac{5 \sqrt{31}}{9}, r=\frac{4 \sqrt{31}}{3 \sqrt{15}}, R=\frac{4 \sqrt{155}}{9 \sqrt{3}}$.
Solution. We apply the tangent-secant theorem three times:
\[
\begin{gathered}
H F^{2}=H D \cdot H E=\frac{1}{9} \cdot \frac{31}{9}=\frac{31}{81} \Rightarrow H F=\frac{\sqrt{31}}{9} \\
H F^{2}=H C \cdot H B \Rig... | HP=\frac{5\sqrt{31}}{9},r=\frac{4\sqrt{31}}{3\sqrt{15}},R=\frac{4\sqrt{155}}{9\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,434 |
5. Solve the inequality
$$
\left(\log _{\frac{2}{3} x^{2}+\frac{1}{2} x+\frac{5}{6}}\left(1+x^{2}\right) \cdot \log _{\frac{2}{3} x^{2}+\frac{1}{2} x+\frac{5}{6}}\left(1-x^{2}\right)+1\right) \log _{1-x^{4}}\left(\frac{2 x^{2}}{3}+\frac{x}{2}+\frac{5}{6}\right) \geqslant 1
$$ | Answer: $x \in\left[-\frac{1}{2} ; 0\right) \cup\left(0 ; \frac{1}{5}\right] \cup\left(\frac{1}{4} ; \frac{1}{2}\right]$.
Solution. The domain of the logarithms in the inequality is determined by the conditions
$$
\begin{gathered}
1-x^{2}>0 \Leftrightarrow-10 \Leftrightarrow x \in \mathbb{R} \\
1+x^{2} \neq 1 \Leftri... | x\in[-\frac{1}{2};0)\cup(0;\frac{1}{5}]\cup(\frac{1}{4};\frac{1}{2}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,435 |
6. A circle, the center of which lies on the line $y=b$, intersects the parabola $y=\frac{5}{12} x^{2}$ at least at three points; one of these points is the origin, and two of the remaining points lie on the line $y=\frac{5}{12} x+b$. Find all values of $b$ for which the described configuration is possible. | Answer: $b=\frac{169}{60}$.
Solution. First, consider $b>0$. Let the origin be denoted as $O(0 ; 0)$, the center of the circle as $Q(a ; b)$ (since it lies on the line $y=b$, its ordinate is $b$); the points of intersection of the line with the parabola as $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)... | \frac{169}{60} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,436 |
7. On the edges $A C, B C, B S, A S$ of a regular triangular pyramid $S A B C$ with vertex $S$, points $K, L, M, N$ are chosen respectively. It is known that points $K, L, M, N$ lie in the same plane, and $K L = M N = 2, K N = L M = 9$. In the quadrilateral $K L M N$, there are two circles $\Omega_{1}$ and $\Omega_{2}$... | Answer: a) $\angle S A B=\arccos \frac{2}{3} ;$ b) $C Q=\frac{7}{3}$.
Solution. The opposite sides of quadrilateral $K L M N$ are equal in pairs, so it is a parallelogram. Since the plane $(K L M N)$ intersects the planes $(A B C)$ and $(A B S)$ along parallel lines $K L$ and $M N$, these lines are parallel to the lin... | \angleSAB=\arccos\frac{2}{3};CQ=\frac{7}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,437 |
1. Given quadratic trinomials $f_{1}(x)=x^{2}+a x+3, f_{2}(x)=x^{2}+2 x-b, f_{3}(x)=x^{2}+2(a-1) x+b+6$ and $f_{4}(x)=x^{2}+(4-a) x-2 b-3$. Let the differences of their roots be $A, B, C$ and $D$, respectively, and given that $|A| \neq|B|$. Find the ratio $\frac{C^{2}-D^{2}}{A^{2}-B^{2}}$. The values of $A, B, C, D, a,... | Answer: 3.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=$ $\frac{\sqrt{T}}{|a|} \cdot$ Applying t... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,438 |
2. It is known that $\frac{\cos x-\sin x}{\cos y}=2 \sqrt{2} \operatorname{ctg} \frac{x+y}{2}$ and $\frac{\sin x+\cos x}{\sin y}=\frac{1}{\sqrt{2}} \operatorname{tg} \frac{x+y}{2}$. Find all possible values of the expression $\operatorname{tg}(x+y)$, given that there are at least three. | Answer: $1, \frac{4}{3}$ or $-\frac{4}{3}$.
Solution. Multiplying the two given equations, we get $\frac{\cos ^{2} x-\sin ^{2} x}{\cos y \sin y}=2$, which is equivalent to the following on the domain of definition:
$$
\cos 2 x=2 \sin y \cos y \Leftrightarrow \cos 2 x+\cos \left(2 y+\frac{\pi}{2}\right)=0 \Leftrightar... | 1,\frac{4}{3},-\frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,439 |
3. On the table, there are 200 different cards with numbers $201, 203, 205, \ldots, 597, 599$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the sum of the numbers on the selected cards is divisible by 3? | Answer: 437844.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 2. Therefore, the remainders of these numbers when divided by 3 alternate. Indeed, if one of these numbers is divisible by 3, i.e., has the form $3k$, where $k \in \mathbb{N}$, then the next nu... | 437844 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,440 |
4. Circles $\Omega$ and $\omega$ touch externally at point $F$, and their common external tangent touches circles $\Omega$ and $\omega$ at points $A$ and $B$, respectively. Line $\ell$ passes through point $B$, intersects circle $\omega$ again at point $C$, and intersects $\Omega$ at points $D$ and $E$ (point $D$ is be... | Answer: $H P=\frac{5 \sqrt{14}}{2}, r=\frac{3 \sqrt{7}}{\sqrt{10}}, R=\frac{3 \sqrt{35}}{\sqrt{2}}$.
Solution. We apply the tangent-secant theorem three times:
$$
\begin{gathered}
H F^{2}=H C \cdot H B=2 \cdot 7=14 \Rightarrow H F=\sqrt{14} \\
H F^{2}=H D \cdot H E \Rightarrow H E=\frac{H F^{2}}{H D}=7 \\
B A^{2}=B D... | HP=\frac{5\sqrt{14}}{2},r=\frac{3\sqrt{7}}{\sqrt{10}},R=\frac{3\sqrt{35}}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,441 |
5. Solve the inequality
$$
\left(\log _{\frac{1}{6} x^{2}+\frac{x}{3}+\frac{19}{27}}\left(1+\frac{x^{2}}{4}\right) \cdot \log _{\frac{1}{6} x^{2}+\frac{x}{3}+\frac{19}{27}}\left(1-\frac{x^{2}}{4}\right)+1\right) \log _{1-\frac{1}{16} x^{4}}\left(\frac{x^{2}}{6}+\frac{x}{3}+\frac{19}{27}\right) \geqslant 1
$$ | Answer: $x \in\left[-\frac{4}{3} ; 0\right) \cup\left(0 ; \frac{8}{15}\right] \cup\left(\frac{2}{3} ; \frac{4}{3}\right]$.
Solution. The domain of definition of the logarithms in the inequality is determined by the conditions
$$
\begin{gathered}
1-\frac{x^{2}}{4}>0 \Leftrightarrow-20 \Leftrightarrow x \in \mathbb{R},... | x\in[-\frac{4}{3};0)\cup(0;\frac{8}{15}]\cup(\frac{2}{3};\frac{4}{3}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,442 |
6. A circle, the center of which lies on the line $y=b$, intersects the parabola $y=\frac{4}{3} x^{2}$ at least at three points; one of these points is the origin, and two of the remaining points lie on the line $y=\frac{4}{3} x+b$. Find all values of $b$ for which the described configuration is possible. | Answer: $b=\frac{25}{12}$.
Solution. First, consider $b>0$. Let the origin be denoted as $O(0 ; 0)$, the center of the circle as $Q(a ; b)$ (since it lies on the line $y=b$, its ordinate is $b$); the points of intersection of the line with the parabola as $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)\... | \frac{25}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,443 |
7. On the edges $A C, B C, B S, A S$ of a regular triangular pyramid $S A B C$ with vertex $S$, points $K, L, M, N$ are chosen respectively. It is known that points $K, L, M, N$ lie in the same plane, and $K L = M N = 4, K N = L M = 9$. In the quadrilateral $K L M N$, there are two circles $\Omega_{1}$ and $\Omega_{2}$... | Answer: a) $\angle S A B=\arccos \frac{1}{3}$; b) $C Q=\frac{25}{3}$.
Solution. The opposite sides of quadrilateral $K L M N$ are pairwise equal, so it is a parallelogram. Since the plane $(K L M N)$ intersects the planes $(A B C)$ and $(A B S)$ along parallel lines $K L$ and $M N$, these lines are parallel to the lin... | \angleSAB=\arccos\frac{1}{3};CQ=\frac{25}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,444 |
1. Given quadratic trinomials $f_{1}(x)=x^{2}+2 x+a, f_{2}(x)=x^{2}+b x-1, f_{3}(x)=2 x^{2}+(6-b) x+3 a+1$ and $f_{4}(x)=2 x^{2}+(3 b-2) x-a-3$. Let the differences of their roots be $A, B, C$ and $D$, respectively, and given that $|A| \neq|B|$. Find the ratio $\frac{C^{2}-D^{2}}{A^{2}-B^{2}}$. The values of $A, B, C, ... | Answer: 2.
Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=$ $\frac{\sqrt{T}}{|a|}$. Applying this f... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,445 |
2. It is known that $\frac{\cos x-\sin x}{\cos y}=\frac{\sqrt{2}}{3} \operatorname{tg} \frac{x+y}{2}$ and $\frac{\sin x+\cos x}{\sin y}=3 \sqrt{2} \operatorname{ctg} \frac{x+y}{2}$. Find all possible values of the expression $\operatorname{tg}(x+y)$, given that there are at least three. | Answer: $1, \frac{3}{4}$ or $-\frac{3}{4}$.
Solution. Multiplying the two given equations, we get $\frac{\cos ^{2} x-\sin ^{2} x}{\cos y \sin y}=2$, which is equivalent to the following on the domain of definition:
$$
\cos 2 x=2 \sin y \cos y \Leftrightarrow \cos 2 x+\cos \left(2 y+\frac{\pi}{2}\right)=0 \Leftrightar... | 1,\frac{3}{4},-\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,446 |
3. On the table, there are 160 different cards with numbers $5, 10, 15, \ldots, 795, 800$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the sum of the numbers on the selected cards is divisible by 3? | Answer: 223342.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a common difference of 5. Therefore, the remainders when these numbers are divided by 3 alternate. Indeed, if one of these numbers is divisible by 3, i.e., has the form $3k$, where $k \in \mathbb{N}$, then the... | 223342 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,447 |
4. Circles $\Omega$ and $\omega$ touch each other externally at point $F$, and their common external tangent touches circles $\Omega$ and $\omega$ at points $A$ and $B$, respectively. Line $\ell$ passes through point $B$, intersects circle $\omega$ again at point $C$, and intersects $\Omega$ at points $D$ and $E$ (poin... | Answer: $H P=6, r=5 \sqrt{\frac{2}{3}}, R=5 \sqrt{\frac{2}{3}}$.
Solution. We apply the tangent-secant theorem three times:
$$
\begin{gathered}
H F^{2}=H D \cdot H E=\frac{1}{7} \cdot 7=1 \Rightarrow H F=1 \\
H F^{2}=H C \cdot H B \Rightarrow H B=\frac{H F^{2}}{H C}=7 \\
B A^{2}=B D \cdot B E=\frac{50}{7} \cdot 14=10... | HP=6,r=5\sqrt{\frac{2}{3}},R=5\sqrt{\frac{2}{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,448 |
5. Solve the inequality
$$
\left(\log _{\frac{2}{27} x^{2}-\frac{2}{9} x+\frac{19}{27}}\left(1+\frac{x^{2}}{9}\right) \cdot \log _{\frac{2}{27} x^{2}-\frac{2}{9} x+\frac{19}{27}}\left(1-\frac{x^{2}}{9}\right)+1\right) \log _{1-\frac{1}{81} x^{4}}\left(\frac{2 x^{2}}{27}-\frac{2 x}{9}+\frac{19}{27}\right) \geqslant 1
$... | Answer: $x \in[-2 ;-1) \cup\left[-\frac{4}{5} ; 0\right) \cup(0 ; 2]$.
Solution. The domain of definition of the logarithms in the inequality is determined by the conditions
$$
\begin{gathered}
1-\frac{x^{2}}{9}>0 \Leftrightarrow-30 \Leftrightarrow x \in \mathbb{R}, \\
1+\frac{x^{2}}{9} \neq 1 \Leftrightarrow x \neq ... | x\in[-2;-1)\cup[-\frac{4}{5};0)\cup(0;2] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,449 |
6. A circle, the center of which lies on the line $y=b$, intersects the parabola $y=\frac{12}{5} x^{2}$ at least at three points; one of these points is the origin, and two of the remaining points lie on the line $y=\frac{12}{5} x+b$. Find all values of $b$ for which the described configuration is possible. | Answer: $b=\frac{169}{60}$.
Solution. First, consider $b>0$. Let the origin be denoted as $O(0 ; 0)$, the center of the circle as $Q(a ; b)$ (since it lies on the line $y=b$, its ordinate is $b$); the points of intersection of the line with the parabola as $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)... | \frac{169}{60} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,450 |
7. On the edges $A C, B C, B S, A S$ of a regular triangular pyramid $S A B C$ with vertex $S$, points $K, L, M, N$ are chosen respectively. It is known that points $K, L, M, N$ lie in the same plane, and $K L = M N = 14, K N = L M = 25$. In the quadrilateral $K L M N$, there are two circles $\Omega_{1}$ and $\Omega_{2... | Answer: a) $\angle S A B=\arccos \frac{3}{5} ;$ b) $C Q=\frac{77}{5}$.
Solution. The opposite sides of quadrilateral $K L M N$ are pairwise equal, so it is a parallelogram. Since the plane $(K L M N)$ intersects the planes $(A B C)$ and $(A B S)$ along parallel lines $K L$ and $M N$, these lines are parallel to the li... | \angleSAB=\arccos\frac{3}{5};CQ=\frac{77}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,451 |
1. Given a linear function $f(x)$. It is known that the distance between the points of intersection of the graphs $y=x^{2}+1$ and $y=f(x)$ is $3 \sqrt{2}$, and the distance between the points of intersection of the graphs $y=x^{2}$ and $y=f(x)-2$ is $\sqrt{10}$. Find the distance between the points of intersection of t... | Answer: $\sqrt{26}$.
Solution. Let $f(x)=a x+b$. Then the abscissas of the points of intersection of the graphs in the first case are determined from the equation $x^{2}+1=a x+b$, and in the second case - from the equation $x^{2}=a x+b-2$. Consider the first case in more detail. The equation has the form $x^{2}-a x-(b... | \sqrt{26} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,452 |
2. It is known that $\frac{\cos 3 x}{(2 \cos 2 x-1) \cos y}=\frac{2}{5}+\cos ^{2}(x+y)$ and $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin y}=\frac{3}{5}+\sin ^{2}(x+y)$. Find all possible values of the expression $\cos (x+3 y)$, given that there are at least two. | Answer: -1 or $\frac{1}{5}$.
Solution. Note that
$$
\begin{aligned}
& \cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x=\cos 2 x \cos x-2 \sin ^{2} x \cos x= \\
& \quad=\cos 2 x \cos x-(1-\cos 2 x) \cos x=(2 \cos 2 x-1) \cos x \\
& \sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\sin x \cos 2 x= \\
& \quad=2 \sin x \cos ^{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,453 |
3. There are 207 different cards with numbers $1,2,3,2^{2}, 3^{2}, \ldots, 2^{103}, 3^{103}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer divisible by 6? | Answer: 267903.
Solution. Consider the case when one of the selected cards has a one written on it. Then, on the other two cards, even powers of two and three must be recorded. There are 51 ways to choose an even power of two and 51 ways to choose an even power of three, and since this choice is made independently, th... | 267903 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,454 |
4. Chords $AB$ and $CD$ of circle $\Gamma$ with center $O$ have a length of 4. The extensions of segments $BA$ and $CD$ beyond points $A$ and $D$ respectively intersect at point $P$. Line $PO$ intersects segment $AC$ at point $L$, such that $AL: LC = 2: 3$.
a) Find $AP$.
b) Suppose additionally that the radius of cir... | Answer: $A P=8, P T=\frac{\sqrt{409}-5}{2}, S_{\triangle A P C}=\frac{5760}{409}$.
Solution. a) Drop perpendiculars $O H$ and $O N$ from point $O$ to the chords $C D$ and $A B$ respectively. Since these chords are equal, the distances from the center of the circle to them are also equal, so $O H=O N$. Right triangles ... | AP=8,PT=\frac{\sqrt{409}-5}{2},S_{\triangleAPC}=\frac{5760}{409} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,455 |
5. Solve the inequality $\log _{1+x^{2}}\left(1+27 x^{5}\right)+\log _{1-2 x^{2}+27 x^{4}}\left(1+x^{2}\right) \leqslant 1+\log _{1-2 x^{2}+27 x^{4}}\left(1+27 x^{5}\right)$. Answer: $x \in\left(-\frac{1}{\sqrt[5]{27}} ;-\frac{1}{3}\right] \cup\left(-\sqrt{\frac{2}{27}} ; 0\right) \cup\left(0 ; \sqrt{\frac{2}{27}}\righ... | Solution. Let $1+x^{2}=u, 1-2 x^{2}+27 x^{4}=v, 1+27 x^{5}=w$. Then the inequality takes the form $\log _{u} w+\log _{v} u-\log _{v} w-1 \leqslant 0$. It can be transformed as follows:
$$
\begin{aligned}
& \log _{u} w+\frac{1}{\log _{u} v}-\frac{\log _{u} w}{\log _{u} v}-1 \leqslant 0 \Leftrightarrow \frac{\log _{u} v... | x\in(-\frac{1}{\sqrt[5]{27}};-\frac{1}{3}]\cup(-\sqrt{\frac{2}{27}};0)\cup(0;\sqrt{\frac{2}{27}})\cup{\frac{1}{3}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,456 |
6. Find all values of the parameter $a$, for each of which the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=26(y \sin 2 a-x \cos 2 a) \\
x^{2}+y^{2}=26(y \cos 3 a-x \sin 3 a)
\end{array}\right.
$$
has two solutions $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$ such that the distance betw... | Answer: $a=\frac{\pi}{10} \pm \frac{2}{5} \operatorname{arctg} \frac{12}{5}+\frac{2 \pi n}{5}, \frac{\pi}{10}+\frac{2 \pi n}{5}$, where $n \in \mathbb{Z}$.
Solution. By completing the squares, we rewrite the system as
$$
\left\{\begin{array}{l}
(x+13 \cos 2 a)^{2}+(y-13 \sin 2 a)^{2}=169 \\
(x+13 \sin 3 a)^{2}+(y-13 ... | \frac{\pi}{10}\\frac{2}{5}\operatorname{arctg}\frac{12}{5}+\frac{2\pin}{5},\frac{\pi}{10}+\frac{2\pin}{5},n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,457 |
7. Given a truncated pyramid $A B C A_{1} B_{1} C_{1}$ with lateral edges $A A_{1}, B B_{1}, C C_{1}\left(A B C \| A_{1} B_{1} C_{1}\right)$, such that the triangle $A B A_{1}$ is equilateral. On the edge $C C_{1}$, perpendicular to the base $A B C$ of the pyramid, there is a point $M$ such that $C M: M C_{1}=1: 2$. Th... | Answer: a) $AB=\sqrt{15}$; b) $\angle\left(CC_{1}, ABA_{1}\right)=\arcsin \frac{1}{\sqrt{6}}, A_{1}C_{1}=\sqrt{\frac{5}{2}}$.
Solution. a) Let $O$ be the center of the sphere $\Omega$, $Q$ be the center of the triangle $ABA_{1}$, and $A_{1}X$ be its median. Then $O$ lies on the perpendicular $\ell$ to $(ABA_{1})$ pass... | AB=\sqrt{15};\angle(CC_{1},ABA_{1})=\arcsin\frac{1}{\sqrt{6}},A_{1}C_{1}=\sqrt{\frac{5}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,458 |
1. Given a linear function $f(x)$. It is known that the distance between the points of intersection of the graphs $y=x^{2}-1$ and $y=f(x)$ is $\sqrt{34}$, and the distance between the points of intersection of the graphs $y=x^{2}+1$ and $y=f(x)+3$ is $\sqrt{42}$. Find the distance between the points of intersection of ... | Answer: $3 \sqrt{2}$.
Solution. Let $f(x)=a x+b$. Then the abscissas of the points of intersection of the graphs in the first case are determined from the equation $x^{2}-1=a x+b$, and in the second case - from the equation $x^{2}+1=a x+b+3$. Consider the first case in more detail. The equation has the form $x^{2}-a x... | 3\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,459 |
2. It is known that $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin 2 y}=\frac{1}{5}+\cos ^{2}(x-2 y)$ and $\frac{\cos 3 x}{(1-2 \cos 2 x) \cos 2 y}=\frac{4}{5}+\sin ^{2}(x-2 y)$. Find all possible values of the expression $\cos (x-6 y)$, given that there are at least two. Answer: 1 or $-\frac{3}{5}$. | Solution. Note that
$$
\begin{aligned}
& \cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x=\cos 2 x \cos x-2 \sin ^{2} x \cos x= \\
& \quad=\cos 2 x \cos x-(1-\cos 2 x) \cos x=(2 \cos 2 x-1) \cos x \\
& \sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\sin x \cos 2 x= \\
& \quad=2 \sin x \cos ^{2} x+\sin x \cos 2 x=\sin x(1+... | -\frac{3}{5}or1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,460 |
3. There are 183 different cards with numbers $1,2,11,2^{2}, 11^{2}, \ldots, 2^{91}, 11^{91}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the selected cards is a square of an integer divisible by 22? | Answer: 184275.
Solution. Consider the case when one of the selected cards has the number one written on it. Then, on the other two cards, even powers of the numbers two and eleven must be recorded. There are 45 ways to choose an even power of two and 45 ways to choose an even power of eleven, and since this choice is... | 184275 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,461 |
4. Chords $AB$ and $CD$ of circle $\Gamma$ with center $O$ have a length of 3. The extensions of segments $BA$ and $CD$ beyond points $A$ and $D$ respectively intersect at point $P$. Line $PO$ intersects segment $AC$ at point $L$, such that $AL: LC = 1: 3$.
a) Find $AP$.
b) Suppose additionally that the radius of cir... | Answer: $A P=\frac{3}{2}, P T=\sqrt{13}-\frac{5}{2}, S_{\triangle A P C}=\frac{81}{26}$.
Solution. a) Drop perpendiculars $O H$ and $O N$ from point $O$ to the chords $C D$ and $A B$ respectively. Since these chords are equal, the distances from the center of the circle to them are also equal, so $O H=O N$. Right tria... | AP=\frac{3}{2},PT=\sqrt{13}-\frac{5}{2},S_{\triangleAPC}=\frac{81}{26} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,462 |
5. Solve the inequality $\log _{1+x^{2}}\left(1-8 x^{5}\right)+\log _{1-x^{2}+8 x^{4}}\left(1+x^{2}\right) \leqslant 1+\log _{1-x^{2}+8 x^{4}}\left(1-8 x^{5}\right)$. | Answer: $x \in\left\{-\frac{1}{2}\right\} \bigcup\left(-\frac{1}{2 \sqrt{2}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{2}}\right) \cup\left[\frac{1}{2} ; \frac{1}{\sqrt[5]{8}}\right)$.
Solution. Let $1+x^{2}=u, 1-x^{2}+8 x^{4}=v, 1-8 x^{5}=w$. Then the inequality takes the form $\log _{u} w+\log _{v} u-\log _{v} w-1 \... | x\in{-\frac{1}{2}}\bigcup(-\frac{1}{2\sqrt{2}};0)\cup(0;\frac{1}{2\sqrt{2}})\cup[\frac{1}{2};\frac{1}{\sqrt[5]{8}}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,463 |
6. Find all values of the parameter $a$, for each of which the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=10(x \cos 4 a+y \sin 4 a) \\
x^{2}+y^{2}=10(x \sin a+y \cos a)
\end{array}\right.
$$
has two solutions $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$ such that the distance between ... | Answer: $a=\frac{\pi}{10} \pm \frac{2}{5} \operatorname{arctg} \frac{4}{3}+\frac{2 \pi n}{5}, \frac{\pi}{10}+\frac{2 \pi n}{5}$, where $n \in \mathbb{Z}$.
Solution. By completing the squares, we rewrite the system as
$$
\left\{\begin{array}{l}
(x-5 \cos 4 a)^{2}+(y-5 \sin 4 a)^{2}=25 \\
(x-5 \sin a)^{2}+(y-5 \cos a)^... | \frac{\pi}{10}\\frac{2}{5}\operatorname{arctg}\frac{4}{3}+\frac{2\pin}{5},\frac{\pi}{10}+\frac{2\pin}{5},n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,464 |
7. Given a truncated pyramid $A B C A_{1} B_{1} C_{1}$ with lateral edges $A A_{1}, B B_{1}, C C_{1}\left(A B C \| A_{1} B_{1} C_{1}\right)$, such that the triangle $B B_{1} C$ is equilateral. On the edge $A A_{1}$, perpendicular to the base $A B C$ of the pyramid, there is a point $N$ such that $A N: N A_{1}=1: 2$. Th... | Answer: a) $B B_{1}=\sqrt{15}$; b) $\angle\left(A A_{1}, B B_{1} C\right)=\pi / 4, A_{1} B_{1}=2-\sqrt{\frac{3}{2}}$.
Solution. a) Let $O$ be the center of the sphere $\Omega, Q$ be the center of the triangle $B C B_{1}, B_{1} X$ be its median. Then $O$ lies on the perpendicular $\ell$ to $\left(B C B_{1}\right)$ pass... | \frac{)BB_{1}=\sqrt{15};b)\angle(AA_{1},BB_{1}C)=\pi}{4,A_{1}B_{1}=2-\sqrt{\frac{3}{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,465 |
1. Given a linear function $f(x)$. It is known that the distance between the points of intersection of the graphs $y=x^{2}-2$ and $y=f(x)$ is $\sqrt{26}$, and the distance between the points of intersection of the graphs $y=x^{2}$ and $y=f(x)+1$ is $3 \sqrt{2}$. Find the distance between the points of intersection of t... | Answer: $\sqrt{10}$.
Solution. Let $f(x)=a x+b$. Then the abscissas of the points of intersection of the graphs in the first case are determined from the equation $x^{2}-2=a x+b$, and in the second case - from the equation $x^{2}=a x+b+1$. Consider the first case in more detail. The equation has the form $x^{2}-a x-(b... | \sqrt{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,466 |
2. It is known that $\frac{\cos 3 x}{(2 \cos 2 x-1) \cos y}=\frac{2}{3}+\cos ^{2}(x-y)$ and $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin y}=-\frac{1}{3}-\sin ^{2}(x-y)$. Find all possible values of the expression $\cos (x-3 y)$, given that there are at least two. Answer: -1 or $-\frac{1}{3}$. | Solution. Note that
$$
\begin{aligned}
\cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x & =\cos 2 x \cos x-2 \sin ^{2} x \cos x= \\
& =\cos 2 x \cos x-(1-\cos 2 x) \cos x=(2 \cos 2 x-1) \cos x
\end{aligned}
$$
$$
\begin{aligned}
& \sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\sin x \cos 2 x= \\
& \quad=2 \sin x \cos ^{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,467 |
3. There are 195 different cards with numbers $1, 5, 7, 5^{2}, 7^{2}, \ldots, 5^{97}, 7^{97}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer divisible by 35? | Answer: 223488.
Solution. Consider the case when one of the selected cards has a one written on it. Then, on the other two cards, even powers of fives and sevens must be recorded. There are 48 ways to choose an even power of five and 48 ways to choose an even power of seven, and since this choice is made independently... | 223488 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,468 |
4. Chords $AB$ and $CD$ of circle $\Gamma$ with center $O$ have a length of 5. The extensions of segments $BA$ and $CD$ beyond points $A$ and $D$ respectively intersect at point $P$. Line $PO$ intersects segment $AC$ at point $L$, such that $AL: LC = 1: 2$.
a) Find $AP$.
b) Suppose additionally that the radius of cir... | Answer: $A P=5, P T=\frac{3 \sqrt{41}-13}{2}, S_{\triangle A P C}=\frac{1000}{41}$.
Solution. a) Drop perpendiculars $O H$ and $O N$ from point $O$ to the chords $C D$ and $A B$ respectively. Since these chords are equal, the distances from the center of the circle to them are also equal, so $O H=O N$. Right triangles... | AP=5,PT=\frac{3\sqrt{41}-13}{2},S_{\triangleAPC}=\frac{1000}{41} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,469 |
5. Solve the inequality $\log _{1+x^{2}}\left(1+8 x^{5}\right)+\log _{1-3 x^{2}+16 x^{4}}\left(1+x^{2}\right) \leqslant 1+\log _{1-3 x^{2}+16 x^{4}}\left(1+8 x^{5}\right)$. Answer: $x \in\left(-\frac{1}{\sqrt[3]{8}} ;-\frac{1}{2}\right] \cup\left(-\frac{\sqrt{3}}{4} ; 0\right) \cup\left(0 ; \frac{\sqrt{3}}{4}\right) \c... | Solution. Let $1+x^{2}=u, 1-3 x^{2}+16 x^{4}=v, 1+8 x^{5}=w$. Then the inequality takes the form $\log _{u} w+\log _{v} u-\log _{v} w-1 \leqslant 0$. It can be transformed as follows:
$$
\begin{aligned}
& \log _{u} w+\frac{1}{\log _{u} v}-\frac{\log _{u} w}{\log _{u} v}-1 \leqslant 0 \Leftrightarrow \frac{\log _{u} v ... | x\in(-\frac{1}{\sqrt[3]{8}};-\frac{1}{2}]\cup(-\frac{\sqrt{3}}{4};0)\cup(0;\frac{\sqrt{3}}{4})\cup{\frac{1}{2}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,470 |
6. Find all values of the parameter $a$, for each of which the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=26(y \sin a-x \cos a) \\
x^{2}+y^{2}=26(y \cos 2 a-x \sin 2 a)
\end{array}\right.
$$
has two solutions $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$ such that the distance between ... | Answer: $a=\frac{\pi}{6} \pm \frac{2}{3} \operatorname{arctg} \frac{5}{12}+\frac{2 \pi n}{3}, \frac{\pi}{6}+\frac{2 \pi n}{3}$, where $n \in \mathbb{Z}$.
Solution. By completing the squares, we rewrite the system as
$$
\left\{\begin{array}{l}
(x+13 \cos a)^{2}+(y-13 \sin a)^{2}=169 \\
(x+13 \sin 2 a)^{2}+(y-13 \cos 2... | \frac{\pi}{6}\\frac{2}{3}\operatorname{arctg}\frac{5}{12}+\frac{2\pin}{3},\frac{\pi}{6}+\frac{2\pin}{3},n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,471 |
7. Given a truncated pyramid $A B C A_{1} B_{1} C_{1}$ with lateral edges $A A_{1}, B B_{1}, C C_{1}\left(A B C \| A_{1} B_{1} C_{1}\right)$, such that the triangle $A B A_{1}$ is equilateral. On the edge $C C_{1}$, perpendicular to the base $A B C$ of the pyramid, there is a point $M$ such that $C M: M C_{1}=1: 2$. Th... | Answer: a) $AB=\sqrt{21}$; b) $\angle\left(C_{1} C, A A_{1} B\right)=\arcsin \sqrt{\frac{2}{3}}, A_{1} C_{1}=\frac{7-2 \sqrt{7}}{3}$.
Solution. a) Let $O$ be the center of the sphere $\Omega, Q$ be the center of the triangle $ABA_{1}, A_{1}X$ be its median. Then $O$ lies on the perpendicular $\ell$ to $(ABA_{1})$ pass... | AB=\sqrt{21};\angle(C_{1}C,AA_{1}B)=\arcsin\sqrt{\frac{2}{3}},A_{1}C_{1}=\frac{7-2\sqrt{7}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,472 |
1. Given a linear function $f(x)$. It is known that the distance between the points of intersection of the graphs $y=x^{2}+2$ and $y=f(x)$ is $\sqrt{10}$, and the distance between the points of intersection of the graphs $y=x^{2}-1$ and $y=f(x)+1$ is $\sqrt{42}$. Find the distance between the points of intersection of ... | Answer: $\sqrt{34}$.
Solution. Let $f(x)=a x+b$. Then the abscissas of the points of intersection of the graphs in the first case are determined from the equation $x^{2}+2=a x+b$, and in the second case - from the equation $x^{2}-1=a x+b+1$. Consider the first case in more detail. The equation has the form $x^{2}-a x-... | \sqrt{34} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,473 |
2. It is known that $\frac{\cos 3 x}{(2 \cos 2 x-1) \cos 2 y}=\frac{1}{6}+\sin ^{2}(x+2 y)$ and $\frac{\sin 3 x}{(2 \cos 2 x+1) \sin 2 y}=\frac{5}{6}+\cos ^{2}(x+2 y)$. Find all possible values of the expression $\cos (x+6 y)$, given that there are at least two. Answer: -1 or $-\frac{2}{3}$. | Solution. Note that
$$
\begin{aligned}
& \cos 3 x=\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x=\cos 2 x \cos x-2 \sin ^{2} x \cos x= \\
& \quad=\cos 2 x \cos x-(1-\cos 2 x) \cos x=(2 \cos 2 x-1) \cos x \\
& \sin 3 x=\sin (2 x+x)=\sin 2 x \cos x+\sin x \cos 2 x= \\
& \quad=2 \sin x \cos ^{2} x+\sin x \cos 2 x=\sin x(1+... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,474 |
3. There are 167 different cards with numbers $1, 3, 11, 3^{2}, 11^{2}, \ldots, 3^{83}, 11^{83}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer divisible by 33? | Answer: 139523.
Solution. Consider the case when one of the selected cards has the number one written on it. Then, on the other two cards, even powers of the numbers three and eleven must be recorded. There are 41 ways to choose an even power of three and 41 ways to choose an even power of eleven, and since this choic... | 139523 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 3,475 |
4. Chords $AB$ and $CD$ of circle $\Gamma$ with center $O$ have a length of 8. The extensions of segments $BA$ and $CD$ beyond points $A$ and $D$ respectively intersect at point $P$. Line $PO$ intersects segment $AC$ at point $L$, such that $AL: LC = 1: 5$.
a) Find $AP$.
b) Suppose additionally that the radius of cir... | Answer: $A P=2, P T=3 \sqrt{5}-5, S_{\triangle A P C}=8$.
Solution. a) Drop perpendiculars $O H$ and $O N$ from point $O$ to the chords $C D$ and $A B$ respectively. Since these chords are equal, the distances from the center of the circle to them are also equal, so
$O H=O N$. Right triangles $O P N$ and $O P H$ are e... | AP=2,PT=3\sqrt{5}-5,S_{\triangleAPC}=8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,476 |
5. Solve the inequality $\log _{1+x^{2}}\left(1-27 x^{5}\right)+\log _{1-3 x^{2}+36 x^{4}}\left(1+x^{2}\right) \leqslant 1+\log _{1-3 x^{2}+36 x^{4}}\left(1-27 x^{5}\right)$. Answer: $x \in\left\{-\frac{1}{3}\right\} \bigcup\left(-\frac{1}{2 \sqrt{3}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{3}}\right) \cup\left[\frac... | Solution. Let $1+x^{2}=u, 1-3 x^{2}+36 x^{4}=v, 1-27 x^{5}=w$. Then the inequality takes the form $\log _{u} w+\log _{v} u-\log _{v} w-1 \leqslant 0$. It can be transformed as follows:
$$
\begin{aligned}
& \log _{u} w+\frac{1}{\log _{u} v}-\frac{\log _{u} w}{\log _{u} v}-1 \leqslant 0 \Leftrightarrow \frac{\log _{u} v... | x\in{-\frac{1}{3}}\bigcup(-\frac{1}{2\sqrt{3}};0)\cup(0;\frac{1}{2\sqrt{3}})\cup[\frac{1}{3};\frac{1}{\sqrt[5]{27}}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 3,477 |
6. Find all values of the parameter $a$, for each of which the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=10(x \cos a+y \sin a) \\
x^{2}+y^{2}=10(x \sin 3 a+y \cos 3 a)
\end{array}\right.
$$
has two solutions $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$ such that the distance between ... | Answer: $a=\frac{\pi}{8} \pm \frac{1}{2} \operatorname{arctg} \frac{3}{4}+\frac{\pi n}{2}, \frac{\pi}{8}+\frac{\pi n}{2}$, where $n \in \mathbb{Z}$.
Solution. By completing the squares, we rewrite the system as
$$
\left\{\begin{array}{l}
(x-5 \cos a)^{2}+(y-5 \sin a)^{2}=25 \\
(x-5 \sin 3 a)^{2}+(y-5 \cos 3 a)^{2}=25... | \frac{\pi}{8}\\frac{1}{2}\operatorname{arctg}\frac{3}{4}+\frac{\pin}{2},\frac{\pi}{8}+\frac{\pin}{2},n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,478 |
7. Given a truncated pyramid $A B C A_{1} B_{1} C_{1}$ with lateral edges $A A_{1}, B B_{1}, C C_{1}\left(A B C \| A_{1} B_{1} C_{1}\right)$, such that the triangle $B B_{1} C$ is equilateral. On the edge $A A_{1}$, perpendicular to the base $A B C$ of the pyramid, there is a point $N$ such that $A N: N A_{1}=1: 2$. Th... | Answer: a) $B B_{1}=\sqrt{21} ;$ b) $\angle\left(A A_{1}, B B_{1} C\right)=\pi / 6, A_{1} B_{1}=\sqrt{6}-\frac{\sqrt{3}}{2}$.
Solution. a) Let $O$ be the center of the sphere $\Omega, Q$ be the center of the triangle $B C B_{1}, B_{1} X$ be its median. Then $O$ lies on the perpendicular $\ell$ to $\left(B C B_{1}\righ... | \frac{)BB_{1}=\sqrt{21};b)\angle(AA_{1},BB_{1}C)=\pi}{6,A_{1}B_{1}=\sqrt{6}-\frac{\sqrt{3}}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,479 |
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x^{2}}\left(x^{2}-3 x+2\right)$, $\log _{x^{2}} \frac{x^{2}}{x-2}$, and $\log _{x^{2}} \frac{x^{2}}{x-1}$ is equal to the sum of the other two. | Answer: $x=3$.
Solution. Note that on the domain of definition, the sum of all three logarithms is
$$
\log _{x^{2}}\left(\frac{x^{2}}{x-2} \cdot \frac{x^{2}}{x-1}\left(x^{2}-3 x+2\right)\right)=\log _{x^{2}} x^{4}=2
$$
Let the number that is equal to the sum of the other two be denoted by $c$, and the two remaining ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,480 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $3(g(x))^{2}+$ $2 f(x)$, if the minimum value of the function $3(f(x))^{2}+2 g(x)$ is $-\frac{19}{6}$. | Answer: $\frac{5}{2}$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=3(f(x))^{2}+$ $2 g(x)$. Expanding the brackets, we get $h(x)=3(a x+b)^{2}+2(a x+c)=3 a^{2} x^{2}+2 a(3 b+1) x+3 b^{2}+$ 2c. The graph of $y=h(x)$ is a parabola opening upwards, and the minimum value is attained at the verte... | \frac{5}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,481 |
3. On each of the lines $y=1$ and $y=6$, there are 200 points with abscissas $1,2,3, \ldots, 200$. In how many ways can three points be chosen from the 400 marked points so that they form the vertices of a right triangle? | Answer: $C_{200}^{2} \cdot 4 + 190 \cdot 2 + 174 \cdot 4 = 80676$.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle lies on the other line. Let $A B C$ be the given triangle with a right angle at vertex $C$, and $C H$ be its height dr... | 80676 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,482 |
4. The numbers $x$ and $y$ are such that the equalities $\operatorname{ctg} x+\operatorname{ctg} y=3$ and $2 \sin (2 x+2 y)=\sin 2 x \sin 2 y$ hold. Find $\operatorname{ctg} x \operatorname{ctg} y$. | Answer: $\frac{3}{2}$.
Solution. On the domain of definition, the first equality is equivalent to the following:
$$
\frac{\cos x \sin y+\cos y \sin x}{\sin x \sin y}=3 \Leftrightarrow \sin (x+y)=3 \sin x \sin y
$$
Transform the second equality:
$$
4 \sin (x+y) \cos (x+y)=4 \sin x \sin y \cos x \cos y
$$
Substitute... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,483 |
5. Circle $\Omega$ with radius $\sqrt{3}$ touches sides $B C$ and $A C$ of triangle $A B C$ at points $K$ and $L$ respectively, and intersects side $A B$ at points $M$ and $N$ ( $M$ lies between $A$ and $N$ ) such that segment $M K$ is parallel to $A C$, $K C=1$, $A L=4$. Find $\angle A C B$, the lengths of segments $M... | Answer: $\angle A C B=\frac{2 \pi}{3}, M K=3, A B=\frac{5 \sqrt{7}}{2}, S_{\triangle C M N}=\frac{45 \sqrt{3}}{28}$.
Solution. Let the center of the circle be denoted as $O$, the foot of the altitude from vertex $C$ of the triangle as $H$, and the angle $B A C$ as $\alpha$.
Since the radius drawn to the point of tang... | \angleACB=\frac{2\pi}{3},MK=3,AB=\frac{5\sqrt{7}}{2},S_{\triangleCMN}=\frac{45\sqrt{3}}{28} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,484 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from seven consecutive natural numbers to some number $a$ is 609, and the sum of the distances from these same seven numbers to some number $b$ is 721. Find all possible values of $a$, given that... | Answer: $a=104, a=191, a=1$.
Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+6$. Notice that if some number lies on the interval $[k ; k+6]$, then the sum of the distances from it to the given seven numbers does not exceed $\frac{7}{2} \cdot 6=21$ (the sum of the distances to the t... | =104,=191,=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,485 |
7. Edge $A_{1} A$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ is perpendicular to its face $A B C D$. Sphere $\Omega$ touches the edges $B B_{1}, B_{1} C_{1}, C_{1} C, C B, C D$, and touches edge $C D$ at a point $K$ such that $C K=9, K D=1$.
a) Find the length of edge $A_{1} A$.
b) Suppose it is addition... | Answer: $A_{1} A=18, V=1944, R=3 \sqrt{10}$.
Solution. a) From the condition, it follows that $B B_{1} \perp A B C D$, therefore $B B_{1} \perp B C$ and the face $B C C_{1} B_{1}$ is a rectangle. A circle (the section of the sphere by the plane $B C C_{1} B_{1}$) can be inscribed in this rectangle, which means this fa... | A_{1}A=18,V=1944,R=3\sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,486 |
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x^{2}}\left(x^{2}-7 x+12\right)$, $\log _{x^{2}} \frac{x^{2}}{x-3}$, and $\log _{x^{2}} \frac{x^{2}}{x-4}$ is equal to the sum of the other two. | Answer: $x=5$.
Solution. Note that on the domain of definition, the sum of all three logarithms is
$$
\log _{x^{2}}\left(\frac{x^{2}}{x-3} \cdot \frac{x^{2}}{x-4}\left(x^{2}-7 x+12\right)\right)=\log _{x^{2}} x^{4}=2
$$
Let the number that is equal to the sum of the other two be denoted by $c$, and the two remaining... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,487 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $2(g(x))^{2}-$ $f(x)$, if the minimum value of the function $2(f(x))^{2}-g(x)$ is $\frac{7}{2}$. | Answer: $-\frac{15}{4}$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=2(f(x))^{2}-g(x)$. Expanding the brackets, we get $h(x)=2(a x+b)^{2}-(a x+c)=2 a^{2} x^{2}+a(4 b-1) x+2 b^{2}-c$. The graph of $y=$ $h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. Th... | -\frac{15}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,488 |
3. On each of the lines $x=2$ and $x=9$, there are 400 points with ordinates $1,2,3, \ldots, 400$. In how many ways can three points be chosen from the 800 marked points so that they form the vertices of a right triangle? | Answer: 321372.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of th... | 321372 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,489 |
4. The numbers $x$ and $y$ are such that the equalities $\operatorname{tg} x - \operatorname{tg} y = 7$ and $2 \sin (2 x - 2 y) = \sin 2 x \sin 2 y$ hold. Find $\operatorname{tg} x \operatorname{tg} y$. | Answer: $-\frac{7}{6}$.
Solution. On the domain of definition, the first equality is equivalent to the following:
$$
\frac{\sin x \cos y-\sin y \cos x}{\cos x \cos y}=3 \Leftrightarrow \sin (x-y)=7 \cos x \cos y
$$
Transform the second equality:
$$
4 \sin (x-y) \cos (x-y)=4 \sin x \sin y \cos x \cos y
$$
Substitut... | -\frac{7}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,490 |
5. Circle G with radius $2 \sqrt{3}$ touches sides $B C$ and $A C$ of triangle $A B C$ at points $K$ and $L$ respectively and intersects side $A B$ at points $M$ and $N$ ( $M$ lies between $A$ and $N$ ) such that segment $M K$ is parallel to $A C, C L=2, B K=3$. Find $\angle A C B$, the lengths of segments $M K, A B$ a... | Answer: $\angle A C B=\frac{2 \pi}{3}, M K=6, A B=5 \sqrt{7}, S_{\triangle B K N}=\frac{9 \sqrt{3}}{14}$.
Solution. Let the center of the circle be denoted as $O$, and the angle $C B A$ as $\beta$.
Since the radius drawn to the point of tangency is perpendicular to the tangent, $\operatorname{to} \operatorname{tg} \a... | \angleACB=\frac{2\pi}{3},MK=6,AB=5\sqrt{7},S_{\triangleBKN}=\frac{9\sqrt{3}}{14} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,491 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from thirty-three consecutive natural numbers to some number $a$ is 3168, and the sum of the distances from these same thirty-three numbers to some number $b$ is 924. Find all possible values of ... | Answer: $a=26, a=-2, a=122$.
Solution. Let the given sequence of consecutive natural numbers be denoted as $k, k+1, \ldots, k+32$. Note that if some number lies on the interval $[k ; k+32]$, then the sum of the distances from it to these thirty-three numbers does not exceed $\frac{33}{2} \cdot 32=528$ (the sum of the ... | =26,=-2,=122 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,492 |
7. Edge $A_{1} A$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ is perpendicular to its face $A B C D$. Sphere $\Omega$ touches the edges $B B_{1}, B_{1} C_{1}, C_{1} C, C B, C_{1} D_{1}$, and touches the edge $C_{1} D_{1}$ at a point $K$ such that $C_{1} K=9, K D_{1}=4$.
a) Find the length of edge $A_{1} A$... | Answer: $A_{1} A=18, V=3888, R=3 \sqrt{13}$.
Solution. a) From the condition, it follows that $B B_{1} \perp A B C D$, therefore $B B_{1} \perp B C$ and the face $B C C_{1} B_{1}$ is a rectangle. A circle (the section of the sphere by the plane $B C C_{1} B_{1}$) can be inscribed in this rectangle, which means this fa... | A_{1}A=18,V=3888,R=3\sqrt{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,493 |
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x^{2}}\left(x^{2}-10 x+21\right)$, $\log _{x^{2}} \frac{x^{2}}{x-7}$, and $\log _{x^{2}} \frac{x^{2}}{x-3}$ is equal to the sum of the other two. | Answer: $x=8$.
Solution. Note that on the domain of definition, the sum of all three logarithms is
$$
\log _{x^{2}}\left(\frac{x^{2}}{x-3} \cdot \frac{x^{2}}{x-7}\left(x^{2}-10 x+21\right)\right)=\log _{x^{2}} x^{4}=2
$$
Let the number that is equal to the sum of the other two be denoted by $c$, and the two remainin... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,494 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $(g(x))^{2}+$ $5 f(x)$, if the minimum value of the function $(f(x))^{2}+5 g(x)$ is -17. | Answer: $\frac{9}{2}$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}+5 g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}+5(a x+c)=a^{2} x^{2}+a(2 b+5) x+b^{2}+5 c$. The graph of $y=$ $h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x... | \frac{9}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,495 |
3. On each of the lines $y=1$ and $y=12$, there are 200 points with abscissas $1,2,3, \ldots, 200$. In how many ways can three points be chosen from the 400 marked points so that they form the vertices of a right triangle? | Answer: 80268.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of the... | 80268 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,496 |
4. The numbers $x$ and $y$ are such that the equalities $\operatorname{ctg} x - \operatorname{ctg} y = 2$ and $5 \sin (2 x - 2 y) = \sin 2 x \sin 2 y$ hold. Find $\operatorname{tg} x \operatorname{tg} y$. | Answer: $-\frac{6}{5}$.
Solution. On the domain of definition, the first equality is equivalent to the following:
$$
\frac{\cos x \sin y-\cos y \sin x}{\sin x \sin y}=3 \Leftrightarrow-\sin (x-y)=2 \sin x \sin y
$$
Transform the second equality:
$$
10 \sin (x-y) \cos (x-y)=4 \sin x \sin y \cos x \cos y
$$
Substitu... | -\frac{6}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,497 |
5. Circle $\Omega$ with radius $\sqrt{3}$ touches sides $B C$ and $A C$ of triangle $A B C$ at points $K$ and $L$ respectively, and intersects side $A B$ at points $M$ and $N$ ( $M$ lies between $A$ and $N$ ) such that segment $M K$ is parallel to $A C, K C=1, A L=6$. Find $\angle A C B$, the lengths of segments $M K, ... | Answer: $\angle A C B=\frac{2 \pi}{3}, M K=3, A B=\frac{7 \sqrt{21}}{4}, S_{\triangle C M N}=\frac{5 \sqrt{3}}{4}$.
Solution. Let the center of the circle be denoted as $O$, the foot of the altitude from vertex $C$ of the triangle as $H$, and the angle $B A C$ as $\alpha$.
Since the radius drawn to the point of tange... | \angleACB=\frac{2\pi}{3},MK=3,AB=\frac{7\sqrt{21}}{4},S_{\triangleCMN}=\frac{5\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,498 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from eleven consecutive natural numbers to some number $a$ is 902, and the sum of the distances from these same eleven numbers to some number $b$ is 374. Find all possible values of $a$, given th... | Answer: $a=107, a=-9, a=25$.
Solution. Let the given sequence of natural numbers be denoted as $k, k+1, \ldots, k+10$. Notice that if some number lies on the interval $[k ; k+10]$, then the sum of the distances from it to these eleven numbers does not exceed $\frac{11}{2} \cdot 10=55$ (the sum of the distances to the ... | =107,=-9,=25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,499 |
7. Edge $A_{1} A$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ is perpendicular to its face $A B C D$. Sphere $\Omega$ touches the edges $B B_{1}, B_{1} C_{1}, C_{1} C, C B, C D$, and touches edge $C D$ at a point $K$ such that $C K=4, K D=1$.
a) Find the length of edge $A_{1} A$.
b) Suppose it is addition... | Answer: $A_{1} A=8, V=256, R=2 \sqrt{5}$.
Solution. a) From the condition, it follows that $B B_{1} \perp A B C D$, therefore $B B_{1} \perp B C$ and the face $B C C_{1} B_{1}$ is a rectangle. A circle can be inscribed in this rectangle (the section of the sphere by the plane $B C C_{1} B_{1}$), which means this face ... | A_{1}A=8,V=256,R=2\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,500 |
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x^{2}}\left(x^{2}-7 x+10\right)$, $\log _{x^{2}} \frac{x^{2}}{x-2}$, and $\log _{x^{2}} \frac{x^{2}}{x-5}$ is equal to the sum of the other two. | Answer: $x=6$.
Solution. Note that on the domain of definition, the sum of all three logarithms is
$$
\log _{x^{2}}\left(\frac{x^{2}}{x-2} \cdot \frac{x^{2}}{x-5}\left(x^{2}-7 x+10\right)\right)=\log _{x^{2}} x^{4}=2
$$
Let the number that is equal to the sum of the other two be denoted by $c$, and the two remaining... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,501 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $(g(x))^{2}-$ $3 f(x)$, if the minimum value of the function $(f(x))^{2}-3 g(x)$ is $\frac{11}{2}$. | Answer: -10.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}-3 g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}-3(a x+c)=a^{2} x^{2}+a(2 b-3) x+b^{2}-3 c$. The graph of $y=$ $h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x-coordinat... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,502 |
3. On each of the lines $x=2$ and $x=15$, there are 400 points with ordinates $1,2,3, \ldots, 400$. In how many ways can three points be chosen from the 800 marked points so that they form the vertices of a right triangle? | Answer: 320868.
Solution. There are two possibilities.
1) The hypotenuse of the triangle lies on one of the lines, and the vertex of the right angle is on the second line. Let $ABC$ be the given triangle with a right angle at vertex $C$, and $CH$ be its height dropped to the hypotenuse. From the proportionality of se... | 320868 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 3,503 |
4. The numbers $x$ and $y$ are such that the equalities $\operatorname{tg} x+\operatorname{tg} y=4$ and $3 \sin (2 x+2 y)=\sin 2 x \sin 2 y$ hold. Find $\operatorname{ctg} x \operatorname{ctg} y$. | Answer: $\frac{7}{6}$.
Solution. On the domain of definition, the first equality is equivalent to the following:
$$
\frac{\sin x \cos y+\sin y \cos x}{\cos x \cos y}=4 \Leftrightarrow \sin (x+y)=4 \cos x \cos y
$$
Transform the second equality:
$$
6 \sin (x+y) \cos (x+y)=4 \sin x \sin y \cos x \cos y
$$
Substitute... | \frac{7}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,504 |
5. Circle G with radius $4 \sqrt{3}$ touches sides $B C$ and $A C$ of triangle $A B C$ at points $K$ and $L$ respectively and intersects side $A B$ at points $M$ and $N$ ( $M$ lies between $A$ and $N$ ) such that segment $M K$ is parallel to $A C, C L=4, B K=3$. Find $\angle A C B$, the lengths of segments $M K, A B$ a... | Answer: $\angle A C B=\frac{2 \pi}{3}, M K=12, A B=7 \sqrt{21}, S_{\triangle B K N}=\frac{3 \sqrt{3}}{7}$.
Solution. Let the center of the circle be denoted as $O$, and the angle $C B A$ as $\beta$.
Since the radius drawn to the point of tangency is perpendicular to the tangent, $\operatorname{to} \operatorname{tg} \... | \angleACB=\frac{2\pi}{3},MK=12,AB=7\sqrt{21},S_{\triangleBKN}=\frac{3\sqrt{3}}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,505 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from twenty-nine consecutive natural numbers to some number $a$ is 1624, and the sum of the distances from these same twenty-nine numbers to some number $b$ is 1305. Find all possible values of $... | Answer: $a=52, a=108, a=7$.
Solution. Let the given sequence of natural numbers be denoted as $k, k+1, \ldots, k+28$. Notice that if some number lies on the interval $[k ; k+28]$, then the sum of the distances from it to the given twenty-nine numbers does not exceed $\frac{29}{2} \cdot 28=406$ (the sum of the distance... | =52,=108,=7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,506 |
7. Edge $A_{1} A$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ is perpendicular to its face $A B C D$. Sphere $\Omega$ touches the edges $B B_{1}, B_{1} C_{1}, C_{1} C, C B, C_{1} D_{1}$, and touches the edge $C_{1} D_{1}$ at a point $K$ such that $C_{1} K=16, K D_{1}=1$.
a) Find the length of edge $A_{1} A... | Answer: $A_{1} A=32, V=8192, R=4 \sqrt{17}$.
Solution. a) From the condition, it follows that $B B_{1} \perp A B C D$, therefore $B B_{1} \perp B C$ and the face $B C C_{1} B_{1}$ is a rectangle. A circle can be inscribed in this rectangle (the section of the sphere by the plane $B C C_{1} B_{1}$), which means this fa... | A_{1}A=32,V=8192,R=4\sqrt{17} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,507 |
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x}\left(x-\frac{3}{2}\right)$, $\log _{x-\frac{3}{2}}(x-3)$, and $\log _{x-3} x$ is equal to the product of the other two. | Answer: $x=\frac{7}{2}, x=\frac{3+\sqrt{13}}{2}$.
Solution. Note that on the domain of definition, the product of all three logarithms is
$$
\log _{x}\left(x-\frac{3}{2}\right) \cdot \log _{x-\frac{3}{2}}(x-3) \cdot \log _{x-3} x=1
$$
Let the number that is the product of the other two be denoted by $c$, and the two... | \frac{7}{2},\frac{3+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,508 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. It is known that the graph of the function $y=(f(x))^{2}$ touches the graph of the function $y=7 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x))... | Answer: $A=-7, A=0$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. The tangency of the graphs $y=(f(x))^{2}$ and $y=7 g(x)$ is equivalent to the equation $(f(x))^{2}=7 g(x)$ having exactly one solution. We get $(a x+b)^{2}=7(a x+c), a^{2} x^{2}+a(2 b-7) x+b^{2}-7 c=0$. The discriminant of this equation is ... | A=-7,A=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,509 |
3. Find the number of distinct reduced quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have at least one root, all their roots are powers of the number 3 with integer non-negative exponents, and their coefficients in absolute value do not exceed \(27^{47}\). | Answer: 5111.
Solution. Such quadratic trinomials can be represented in the form $\left(x-3^{a}\right)\left(x-3^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a \geqslant b$. Expanding the brackets, we get $x^{2}-\left(3^{a}+3^{b}\right) x+3^{a+b}$. According to... | 5111 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,510 |
4. The numbers $x$ and $y$ are such that the equalities $\cos y + \cos x = \sin 3x$ and $\sin 2y - \sin 2x = \cos 4x - \cos 2x$ hold. What is the smallest value that the sum $\sin y + \sin x$ can take? | Answer: $-1-\sin \frac{3 \pi}{8}=-1-\frac{\sqrt{2+\sqrt{2}}}{2}$.
Solution. Transform the second equality:
$$
\sin 2 y=2 \sin x \cos x-2 \sin 3 x \sin x \Leftrightarrow \sin 2 y=2 \sin x(\cos x-\sin 3 x)
$$
Substituting $-\cos y$ for $\cos x-\sin 3 x$, we get $\sin 2 y=-2 \sin x \cos y, 2 \sin y \cos y+2 \sin x \cos... | -1-\frac{\sqrt{2+\sqrt{2}}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,511 |
5. Given a parallelogram $A B C D$. A circle $\Omega$ with radius 5 is circumscribed around triangle $A B M$, where $M$ is the intersection point of the diagonals of the given parallelogram. $\Omega$ intersects the ray $C B$ and the segment $A D$ at points $E$ and $K$ respectively. The length of arc $A E$ is twice the ... | Answer: $A D=10, B K=\frac{48}{5}, P_{E B M}=\frac{84}{5}$.
Solution. Let the degree measures of arcs $B M$ and $A E$ be $2 \alpha$ and $4 \alpha$ respectively. Then, by the inscribed angle theorem, $\angle A B E=\frac{1}{2} \cdot 4 \alpha=2 \alpha, \angle B A M=\frac{1}{2} \cdot 2 \alpha=\alpha$. Therefore, $\angle A... | AD=10,BK=\frac{48}{5},P_{EBM}=\frac{84}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,512 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from sixteen consecutive natural numbers to some number $a$ is 636, and the sum of the distances from these same sixteen numbers to the number $a^{2}$ is 591. Find all possible values of $a$. | Answer: $a=-\frac{5}{4}$.
Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+15$. Note that if some number lies on the segment $[k ; k+15]$, then the sum of the distances from it to the given sixteen numbers does not exceed $8 \cdot 15=120$ (the sum of the distances to the two extreme... | -\frac{5}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,513 |
7. On the edge $B C$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, a point $M$ is chosen. A sphere constructed on the segment $C_{1} M$ as a diameter touches the planes of four faces of the parallelepiped, and one of them at a point lying on the edge $B_{1} B$. It is known that $B M=1, C M=24$. Find the leng... | Answer: $A A_{1}=26, R=5, V=1250$.
Solution. Since the center of the sphere is the midpoint of the segment $C_{1} M$ and lies in the face $B C C_{1} B_{1}$, the sphere does not touch this face. Note that if the segment $C_{1} M$ is not perpendicular to the planes $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$, then the spher... | AA_{1}=26,R=5,V=1250 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,514 |
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x}\left(x-\frac{5}{2}\right)$, $\log _{x-\frac{5}{2}}(x-4)$, and $\log _{x-4} x$ is equal to the product of the other two. | Answer: $x=\frac{9}{2}, x=2+\sqrt{5}$.
Solution. Note that on the domain of definition, the product of all three logarithms is
$$
\log _{x}\left(x-\frac{3}{2}\right) \cdot \log _{x-\frac{3}{2}}(x-3) \cdot \log _{x-3} x=1
$$
Let the number that is the product of the other two be denoted by $c$, and the two remaining ... | \frac{9}{2},2+\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,515 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. It is known that the graph of the function $y=(f(x))^{2}$ touches the graph of the function $y=-12 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x... | Answer: $A=12, A=0$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. The tangency of the graphs $y=(f(x))^{2}$ and $y=-12 g(x)$ is equivalent to the equation $(f(x))^{2}=-12 g(x)$ having exactly one solution. We get $(a x+b)^{2}=-12(a x+c), a^{2} x^{2}+2 a(b+6) x+b^{2}+12 c=0$. A quarter of the discriminant ... | A=12,A=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,516 |
3. Find the number of distinct reduced quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have at least one root, all their roots are powers of the number 5 with integer non-negative exponents, and their coefficients in absolute value do not exceed $122^{85}$. | Answer: 16511.
Solution. Such quadratic trinomials can be represented in the form $\left(x-5^{a}\right)\left(x-5^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a \geqslant b$. Expanding the brackets, we get $x^{2}-\left(5^{a}+5^{b}\right) x+5^{a+b}$. According t... | 16511 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,517 |
4. The numbers $x$ and $y$ are such that the equations $\cos y + \sin x + \cos 3x = 0$ and $\sin 2y - \sin 2x = \cos 4x + \cos 2x$ are satisfied. What is the greatest value that the sum $\sin y + \cos x$ can take? | Answer: $1+\sin \frac{3 \pi}{8}=1+\frac{\sqrt{2+\sqrt{2}}}{2}$.
Solution. Transform the second equality:
$$
\sin 2 y=2 \sin x \cos x+2 \cos 3 x \cos x \Leftrightarrow \sin 2 y=2 \cos x(\sin x+\cos 3 x)
$$
Substituting $-\cos y$ for $\sin x+\cos 3 x$, we get $\sin 2 y=-2 \cos x \cos y, 2 \sin y \cos y+2 \cos x \cos y... | 1+\frac{\sqrt{2+\sqrt{2}}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,518 |
5. Given a parallelogram $A B C D$. A circle $\Omega$ with a diameter of 17 is circumscribed around triangle $A B M$, where $M$ is the point of intersection of the diagonals of the given parallelogram. $\Omega$ intersects the ray $C B$ and the segment $A D$ again at points $E$ and $K$ respectively. The length of arc $A... | Answer: $A D=17, B K=\frac{240}{17}, P_{E B M}=\frac{552}{17}$.
Solution. Let the degree measures of arcs $B M$ and $A E$ be $2 \alpha$ and $4 \alpha$ respectively. Then, by the inscribed angle theorem, $\angle A B E=\frac{1}{2} \cdot 4 \alpha=2 \alpha, \angle B A M=\frac{1}{2} \cdot 2 \alpha=\alpha$. Therefore, $\ang... | AD=17,BK=\frac{240}{17},P_{EBM}=\frac{552}{17} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,519 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from twenty-seven consecutive natural numbers to some number $a$ is 1926, and the sum of the distances from these same twenty-seven numbers to the number $a^{2}$ is 1932. Find all possible values... | Answer: $a=\frac{2}{3}$.
Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+26$. Notice that if some number lies on the segment $[k ; k+26]$, then the sum of the distances from it to the given twenty-seven numbers does not exceed $\frac{27}{2} \cdot 26=351$ (the sum of the distances t... | \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,520 |
7. On the edge $B C$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, a point $M$ is chosen. A sphere constructed on the segment $C_{1} M$ as a diameter touches the planes of four faces of the parallelepiped, and one of them at a point lying on the edge $B_{1} B$. It is known that $B M=1, C M=15$. Find the leng... | Answer: $A A_{1}=17, R=4, V=512$.
Solution. Since the center of the sphere is the midpoint of the segment $C_{1} M$ and lies in the face $B C C_{1} B_{1}$, the sphere does not touch this face. Note that if the segment $C_{1} M$ is not perpendicular to the planes $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$, then the sphere... | AA_{1}=17,R=4,V=512 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,521 |
1. Find all values of $x$, for each of which one of the three given numbers $\log _{x}\left(x-\frac{13}{6}\right)$, $\log _{x-\frac{13}{6}}(x-3)$, and $\log _{x-3} x$ is equal to the product of the other two. | Answer: $x=\frac{11}{3}, x=\frac{3+\sqrt{13}}{2}$.
Solution. Note that on the domain of definition, the product of all three logarithms is
$$
\log _{x}\left(x-\frac{13}{6}\right) \cdot \log _{x-\frac{13}{6}}(x-3) \cdot \log _{x-3} x=1
$$
Let the number that is the product of the other two be denoted by $c$, and the ... | \frac{11}{3},\frac{3+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,522 |
2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. It is known that the graph of the function $y=(f(x))^{2}$ touches the graph of the function $y=4 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x))... | Answer: $A=-4, A=0$.
Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. The tangency of the graphs $y=(f(x))^{2}$ and $y=4 g(x)$ is equivalent to the equation $(f(x))^{2}=4 g(x)$ having exactly one solution. We get $(a x+b)^{2}=4(a x+c), a^{2} x^{2}+2 a(b-2) x+b^{2}-4 c=0$. A quarter of the discriminant of this... | A=-4,A=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,523 |
3. Find the number of distinct reduced quadratic trinomials (i.e., with the leading coefficient equal to 1) with integer coefficients such that they have at least one root, all their roots are powers of the number 7 with integer non-negative exponents, and their coefficients in absolute value do not exceed \(49^{68}\). | Answer: 4760.
Solution. Such quadratic trinomials can be represented in the form $\left(x-7^{a}\right)\left(x-7^{b}\right)$, where $a \geqslant 0$, $b \geqslant 0$ are integers. To avoid repetitions, we assume that $a \geqslant b$. Expanding the brackets, we get $x^{2}-\left(7^{a}+7^{b}\right) x+7^{a+b}$. According to... | 4760 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,524 |
4. The numbers $x$ and $y$ are such that the equalities $\sin y + \cos x = \sin 3x$ and $\sin 2y - \sin 2x = \cos 4x - \cos 2x$ hold. What is the smallest value that the sum $\cos y + \sin x$ can take? | Answer: $-1-\sin \frac{3 \pi}{8}=-1-\frac{\sqrt{2+\sqrt{2}}}{2}$.
Solution. Transform the second equality:
$$
\sin 2 y=2 \sin x \cos x-2 \sin 3 x \sin x \Leftrightarrow \sin 2 y=2 \sin x(\cos x-\sin 3 x)
$$
Substituting $-\sin y$ for $\cos x-\sin 3 x$, we get $\sin 2 y=-2 \sin x \sin y, 2 \sin y \cos y + 2 \sin x \s... | -1-\frac{\sqrt{2+\sqrt{2}}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,525 |
5. Given a parallelogram $A B C D$. A circle $\Omega$ with a diameter of 5 is circumscribed around triangle $A B M$, where $M$ is the point of intersection of the diagonals of the given parallelogram. $\Omega$ intersects the ray $C B$ and the segment $A D$ at points $E$ and $K$ respectively. The length of arc $A E$ is ... | Answer: $B C=5, B K=\frac{24}{5}, P_{E B M}=\frac{42}{5}$.
Solution. Let the degree measures of arcs $B M$ and $A E$ be $2 \alpha$ and $4 \alpha$ respectively. Then, by the inscribed angle theorem, $\angle A B E=\frac{1}{2} \cdot 4 \alpha=2 \alpha, \angle B A M=\frac{1}{2} \cdot 2 \alpha=\alpha$. Therefore, $\angle A ... | BC=5,BK=\frac{24}{5},P_{EBM}=\frac{42}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,526 |
6. Let's call the distance between numbers the absolute value of their difference. It is known that the sum of the distances from eighteen consecutive natural numbers to some number $a$ is 1005, and the sum of the distances from these same eighteen numbers to the number $a^{2}$ is 865. Find all possible values of $a$. | Answer: $a=-\frac{7}{3}$.
Solution. Let the given consecutive natural numbers be denoted by $k, k+1, \ldots, k+17$. Note that if some number lies on the segment $[k ; k+17]$, then the sum of the distances from it to the given eighteen numbers does not exceed $9 \cdot 17=153$ (the sum of the distances to the two extrem... | -\frac{7}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,527 |
7. On the edge $B C$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, a point $M$ is chosen. A sphere constructed on the segment $C_{1} M$ as a diameter touches the planes of four faces of the parallelepiped, and one of them at a point lying on the edge $B_{1} B$. It is known that $B M=1, C M=8$. Find the lengt... | Answer: $R=3, A A_{1}=10, V=162$.
Solution. Since the center of the sphere is the midpoint of the segment $C_{1} M$ and lies in the face $B C C_{1} B_{1}$, the sphere does not touch this face. Note that if the segment $C_{1} M$ is not perpendicular to the planes $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$, then the sphere... | R=3,AA_{1}=10,V=162 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,528 |
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