problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
$9.247 \log _{3}\left(4^{x}+1\right)+\log _{4^{x}+1} 3>2.5$ | Solution. Let $y=\log _{3}\left(4^{x}+1\right)$, then $y+\frac{1}{y}>\frac{5}{2} \Leftrightarrow 2 y^{2}-5 y+2>0 \Leftrightarrow$ 200
$\Leftrightarrow\left[\begin{array}{l}y>2, \\ y<\frac{1}{2} .\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}\log _{3}\left(4^{x}+1\right)>2, \\ \log _{3}\left(4^{x}+1\right)<\... | x\in(-\infty;\log_{4}(\sqrt{3}-1))\cup(\frac{3}{2};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,004 |
$9.248 \log _{3}\left(3^{x}-1\right) \log _{1 / 3}\left(3^{x+2}-9\right)>-3$.
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$9.248 \log _{3}\left(3^{x}-1\right) \log _{1 / 3}\left(3^{x+2}-9\right)>-3$. | Solution. $\log _{3}\left(3^{x}-1\right) \cdot \log _{3} 9\left(3^{x}-1\right)<3$. Let $y=\log _{3}\left(3^{x}-1\right)$, then $y(y+2)<3 \Leftrightarrow-3<y<1 \Leftrightarrow \log _{3} \frac{1}{27}<\log _{3}\left(3^{x}-1\right)<\log _{3} 3 \Leftrightarrow$ $\Leftrightarrow \frac{1}{27}<3^{x}-1<3 \Leftrightarrow \frac{2... | x\in(\log_{3}\frac{28}{27};\log_{3}4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,005 |
$9.249 \log _{a} \frac{1+\log _{a}^{2} x}{1-\log _{a} x}<0$.
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$9.249 \log _{a} \frac{1+\log _{a}^{2} x}{1-\log _{a} x}<0$. | Solution. Let $y=\log _{a} x$, then $\log _{a} \frac{1+y^{2}}{1-y}1, \\ \frac{1+y^{2}}{1-y}0\end{array} \Leftrightarrow\left\{\begin{array}{l}a>1, \\ y(y+1)0\end{array} \Leftrightarrow\left\{\begin{array}{l}a>1, \\ -11, \\ \frac{1}{a}1\end{array} \Leftrightarrow\left\{\begin{array}{l}0\frac{1}{a}, \\ a1$, then $x \in\l... | x\in(-\infty;-1)\cup(0;1)\cup(1;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,006 |
$9.251 \quad \frac{x^{2}-|x|-12}{x-3} \geq 2 x$. | Solution. The original inequality is equivalent to the inequality:
$\frac{x^{2}-6 x+|x|+12}{x-3} \leq 0$.
1) $\left\{\begin{array}{l}x \geq 0, \\ \frac{x^{2}-5 x+12}{x-3} \leq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 0, \\ \frac{1}{x-3} \leq 0\end{array} \Leftrightarrow 0 \leq x<3\right.\right.$.
2) ... | x\in(-\infty;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,007 |
$9.252 \log _{x}\left(x^{3}+1\right) \cdot \log _{x+1} x>2$.
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$9.252 \log _{x}\left(x^{3}+1\right) \cdot \log _{x+1} x>2$. | Solution. $\log _{x}\left((x+1)\left(x^{2}-x+1\right)\right) \cdot \log _{x+1} x>2 \Leftrightarrow$
$\Leftrightarrow\left(\log _{x}(x+1)+\log _{x}\left(x^{2}-x+1\right)\right) \log _{x+1} x>2 \Leftrightarrow 1+\log _{x}\left(x^{2}-x+1\right) \log _{x+1} x>2 \Leftrightarrow$
$\Leftrightarrow \frac{\log _{x}\left(x^{2}... | x\in(2;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,008 |
$9.253 \log _{x}(x+1)<\log _{1 / x}(2-x)$. | Solution.
$\log _{x}(x+1)\frac{1}{2-x} \\ 2-x>0\end{array}, \Leftrightarrow\left\{\begin{array}{l}01, \\ x+10\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ \frac{x^{2}-x-1}{x-2}<0\end{array} \Leftrightarrow \frac{1+\sqrt{5}}{2}<x<2\right.\right.$.
Answer: $\quad x \in(0 ; 1) \cup\left(\frac{1+\sqrt{5}}{2} ... | x\in(0;1)\cup(\frac{1+\sqrt{5}}{2};2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,009 |
$9.254 \log _{3} \log _{0,2} \log _{32} \frac{x-1}{x+5}>0$.
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$9.254 \log _{3} \log _{0.2} \log _{32} \frac{x-1}{x+5}>0$. | Solution. The original inequality is equivalent to the following:
$\log _{0.2} \log _{32} \frac{x-1}{x+5}>1 \Leftrightarrow\left\{\begin{array}{l}\log _{32} \frac{x-1}{x+5}>0, \\ \log _{32} \frac{x-1}{x+5}<1, \\ \frac{x-1}{x+5}>0\end{array} \Leftrightarrow x<-11\right.\right.\right.$.
Answer: $\quad x \in(-\infty ;-1... | x\in(-\infty;-11) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,010 |
$9.255 \log _{x}\left(x^{2}+3 x-3\right)>1$.
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$9.255 \log _{x}\left(x^{2}+3 x-3\right)>1$. | ## Solution.
1) $\left\{\begin{array}{l}x>1, \\ x^{2}+3 x-3>x\end{array} \Leftrightarrow x>1\right.$.
2) $\left\{\begin{array}{l}00\end{array} \Leftrightarrow\left\{\begin{array}{c}0\frac{\sqrt{21}-3}{2}, \\ x3+x$.
Solution.
$|x-1|+|x-2|>3+x$
1) $\left\{\begin{array}{l}x3+x\end{array} \Rightarrow x3+x\end{array} \L... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,011 | |
$9.257 \frac{2}{2+\sqrt{4-x^{2}}}+\frac{1}{2-\sqrt{4-x^{2}}}>\frac{1}{x}$. | Solution.
$\frac{6-\sqrt{4-x^{2}}}{4-4+x^{2}}>\frac{1}{x} \Leftrightarrow\left\{\begin{array}{l}6-x-\sqrt{4-x^{2}}>0, \\ x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}6-x \geq 0, \\ 4-x^{2} \geq 0, \\ (6-x)^{2}>4-x^{2}, \\ x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}|x| \leq 2, \\ x \neq 0 .\... | x\in[-2;0)\cup(0;2] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,012 |
$9.258 \frac{\sqrt{x^{2}-16}}{\sqrt{x-3}}+\sqrt{x-3}>\frac{5}{\sqrt{x-3}}$. | Solution.
$$
\frac{\sqrt{x^{2}-16}+(x-3)}{\sqrt{x-3}}>\frac{5}{\sqrt{x-3}} \Leftrightarrow\left\{\begin{array}{l}
\sqrt{x^{2}-16}>8-x \\
x>3
\end{array}\right.
$$
1) $\left\{\begin{array}{l}8-x \geq 0, \\ x^{2}-16>(8-x)^{2}, \\ x>3\end{array} \Leftrightarrow\left\{\begin{array}{l}x \leq 8, \\ x>5, \\ x>3\end{array} \... | x\in(5;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,013 |
$9.259 \sqrt{4-4 x^{3}+x^{6}}>x-\sqrt[3]{2}$. | Solution. $\sqrt{\left(x^{3}-2\right)^{2}}>x-\sqrt[3]{2} \Leftrightarrow\left|x^{3}-2\right|>x-\sqrt[3]{2} \Leftrightarrow$
$\Leftrightarrow\left[\begin{array}{l}(x-\sqrt[3]{2})\left(x^{2}+\sqrt[3]{2} x+\sqrt[3]{4}\right)>x-\sqrt[3]{2}, \\ (x-\sqrt[3]{2})\left(x^{2}+\sqrt[3]{2} x+\sqrt[3]{4}\right)>0, \\ (x-\sqrt[3]{2... | x\in(-\infty;\sqrt[3]{2})\cup(\sqrt[3]{2};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,014 |
$9.260 \sqrt{x^{4}-2 x^{2}+1}>1-x$.
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$9.260 \sqrt{x^{4}-2 x^{2}+1}>1-x$. | Solution.
$$
\sqrt{\left(x^{2}-1\right)^{2}}>1-x \Leftrightarrow\left|1-x^{2}\right|>1-x \Leftrightarrow\left[\begin{array} { l }
{ 1 - x ^ { 2 } > 1 - x , } \\
{ 1 - x ^ { 2 } < -(1 - x) }
\end{array} \Leftrightarrow \left[\begin{array}{l}
0 < x < 1, \\
x < -2 \text{ or } x > 1
\end{array}\right.\right.\right.
$$
A... | x\in(-\infty;-2)\cup(0;1)\cup(1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,015 |
$9.261 \log _{1 / 2} \frac{\left|x^{2}-2 x\right|+4}{|x+2|+x^{2}} \leq 0$ | Solution. The inequality is equivalent to the following:
$$
\begin{aligned}
& \frac{\left|x^{2}-2 x\right|+4}{|x+2|+x^{2}} \geq 1 \Leftrightarrow\left|x^{2}-2 x\right| \geq x^{2}+|x+2|-4 \Leftrightarrow \\
& \Leftrightarrow\left[\begin{array} { l }
{ x ^ { 2 } - 2 x \geq x ^ { 2 } + | x + 2 | - 4 , } \\
{ x ^ { 2 } -... | x\in(-\infty;\frac{1+\sqrt{17}}{4}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,016 |
$9.262 \log _{x^{2}} \frac{2 x}{|x-3|}<\frac{1}{2}$. | ## Solution.
$\frac{1}{2} \log _{|x|} \frac{2 x}{|x-3|}1, \\ \frac{2 x}{|x-3|} \leq x\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ 2 \leq|x-3|, \\ x \neq 3\end{array} \Leftrightarrow x \geq 5\right.\right.$.
2) $\left\{\begin{array}{l}01$.
## Solution.
$\left(4 x^{2}+2 x+1\right)^{x^{2}-x}>\left(4 x^{2}+... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,017 |
$9.268 \log _{|x-4|}\left(2 x^{2}-9 x+4\right)>1$.
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$9.268 \log _{|x-4|}\left(2 x^{2}-9 x+4\right)>1$. | Solution. $\log _{\mid x-4}\left(2 x^{2}-9 x+4\right)>\log _{\mid x-4}|x-4|$.
1) $\left\{\begin{array}{l}|x-4|>1, \\ 2 x^{2}-9 x+4>|x-4|\end{array} \Leftrightarrow\left\{\begin{array}{l}{\left[\begin{array}{l}x5, \\ x-4-2 x^{2}+9 x-4\end{array}\right.}\end{array} \Leftrightarrow\left\{\begin{array}{l}{\left[\begin{arr... | x\in(-\infty;0)\cup(3;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,018 |
$9.269 \frac{1}{\log _{1 / 2} \sqrt{x+3}} \leq \frac{1}{\log _{1 / 2}(x+1)}$.
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$9.269 \frac{1}{\log _{1 / 2} \sqrt{x+3}} \leq \frac{1}{\log _{1 / 2}(x+1)}$. | Solution. Since $x>-1$, then $x+3>2, \sqrt{x+3}>1, \log _{1 / 2} \sqrt{x+3}1, \\ x+1 \geq \sqrt{x+3}\end{array} \Leftrightarrow\left\{\begin{array}{l}x>0, \\ (x+2)(x-1) \geq 0\end{array} \Leftrightarrow x \geq 1\right.\right.$.
2) $\left\{\begin{array}{l}0<x+1<1, . \\ x+1 \leq \sqrt{x+3}\end{array} \Leftrightarrow\left... | x\in(-1;0)\cup[1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,019 |
$9.270 \log _{x} \frac{3}{8-2 x} \geq-2$.
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$9.270 \log _{x} \frac{3}{8-2 x} \geq-2$. | Solution. $\log _{x} \frac{3}{8-2 x} \geq \log _{x} \frac{1}{x^{2}}$.
1) $\left\{\begin{array}{l}x>1, \\ \frac{3}{8-2 x} \geq \frac{1}{x^{2}}\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ \frac{(x+2)\left(x-\frac{4}{3}\right)}{x-4} \leq 0\end{array} \Leftrightarrow \frac{4}{3} \leq x<4\right.\right.$
2) $\... | x\in(0;1)\cup[\frac{4}{3};4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,020 |
$9.271 \log _{1 / 2}(x-3)-\log _{1 / 2}(x+3)-\log _{\frac{x+3}{x-3}} 2>0$.
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$9.271 \log _{1 / 2}(x-3)-\log _{1 / 2}(x+3)-\log _{\frac{x+3}{x-3}} 2>0$. | Solution. The original inequality is equivalent to the system:
$$
\left\{\begin{array} { l }
{ \operatorname { log } _ { 2 } \frac { x + 3 } { x - 3 } - \operatorname { log } _ { \frac { x + 3 } { x - 3 } } 2 > 0 } \\
{ x > 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{\log _{2}^{2} \frac{x+3}{x-3}-1}... | x\in(3;9) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,021 |
$9.2738 \cdot 3^{\sqrt{x}+\sqrt[4]{x}}+9^{\sqrt[4]{x}+1} \geq 9^{\sqrt{x}}$. | Solution. $8 \cdot 3^{\sqrt{x}} \cdot 3^{\sqrt[4]{x}}+9 \cdot 3^{2 \sqrt[4]{x}} \geq 3^{2 \sqrt{x}}$.
Let $v=3^{\sqrt{x}}, w=3^{\sqrt[4]{x}}$, then $9 w^{2}+8 v w-v^{2} \geq 0 \Leftrightarrow$
$(9 w-v)(w+v) \geq 0 \Leftrightarrow 9 w-v \geq 0 \Leftrightarrow 3^{\sqrt[4]{x}+2} \geq 3^{\sqrt{x}} \Leftrightarrow \sqrt{x... | x\in[0;16] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,022 |
$9.275 \frac{x^{2}-7|x|+10}{x^{2}-6 x+9}<0$.
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$9.275 \frac{x^{2}-7|x|+10}{x^{2}-6 x+9}<0$. | ## Solution.
$$
\begin{aligned}
& \frac{x^{2}-7|x|+10}{(x-3)^{2}}<0 \Leftrightarrow\left\{\begin{array} { l }
{ x ^ { 2 } - 7 | x | + 1 0 < 0 , } \\
{ x \neq 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(|x|-2)(|x|-5)<0, \\
x \neq 3
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begi... | x\in(-5;-2)\cup(2;3)\cup(3;5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,023 |
$9.276 \sin 2 x \cdot \sin 3 x-\cos 2 x \cdot \cos 3 x>\sin 10 x$. | Solution. $-\cos (2 x+3 x)>\sin 10 x \Leftrightarrow \cos 5 x+\sin 10 x0, \\ \sin 5 x-\frac{1}{2}\end{array} \Leftrightarrow\right.$
$\Leftrightarrow \frac{\pi}{2}+2 \pi n<5 x<\frac{7 \pi}{... | x\in(-\frac{\pi}{10}+\frac{2\pin}{5};-\frac{\pi}{30}+\frac{2\pin}{5})\cup(\frac{\pi}{10}+\frac{2\pin}{5};\frac{7\pi}{30}+\frac{2\pin}{5}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,024 |
$9.279 \log _{x} 10-0.5 \log _{a} 10>0, \ 0<a<1$. | ## Solution.
$\log _{x} 10>\frac{1}{\log _{10} a^{2}} \Leftrightarrow \frac{\lg a^{2}}{\lg x}\lg x\end{array} \Leftrightarrow\left\{\begin{array}{l}0<x<1, \\ \lg a^{2}<0, \\ 0<x<a^{2}\end{array} \text { or } \left\{\begin{array}{l}x>1, \\ \lg a^{2}>0, \\ x>a^{2}\end{array} \Leftrightarrow x>1\right.\right.$.
Answer: ... | x\in(0;^{2})\cup(1;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,026 |
$9.282 \sqrt{3 x^{2}+5 x+7}-\sqrt{3 x^{2}+5 x+2}>1$. | ## Solution.
$$
\begin{aligned}
& \sqrt{3 x^{2}+5 x+7}>1+\sqrt{3 x^{2}+5 x+2} \Leftrightarrow \\
& \Leftrightarrow 3 x^{2}+5 x+7>1+2 \sqrt{3 x^{2}+5 x+2}+3 x^{2}+5 x+2 \Leftrightarrow
\end{aligned}
$$
$$
\Leftrightarrow \sqrt{3 x^{2}+5 x+2}<2 \Leftrightarrow\left\{\begin{array} { l }
{ 3 x ^ { 2 } + 5 x + 2 < 4 , } ... | x\in(-2;-1]\cup[-\frac{2}{3};\frac{1}{3}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,028 |
$9.283 \log _{x}^{2} \sqrt{5}-\log _{x} 5 \sqrt{5}+\frac{5}{4}<0$.
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$9.283 \log _{x}^{2} \sqrt{5}-\log _{x} 5 \sqrt{5}+\frac{5}{4}<0$. | Solution. The inequality is equivalent to the following:
$$
\begin{aligned}
& \frac{1}{4} \log _{x}^{2} 5-\frac{3}{2} \log _{x} 5+\frac{5}{4} \frac { 1 } { \log _ { 5 } x } , } \\
{ \frac { 1 } { \log _ { 5 } x } > 1 }
\end{aligned} \Leftrightarrow \left\{\begin{array} { l }
{ \frac { 5 \log _ { 5 } x - 1 } { \log _ ... | x\in(\sqrt[5]{5};5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,029 |
$9.285 \operatorname{ctg} x+\operatorname{ctg}\left(x+\frac{\pi}{2}\right)+2 \operatorname{ctg}\left(x+\frac{\pi}{3}\right)>0$. | ## Solution.
$2 \operatorname{ctg}\left(x+\frac{\pi}{3}\right)+\operatorname{ctg} x-\operatorname{tg} x>0 \Leftrightarrow \operatorname{ctg}\left(x+\frac{\pi}{3}\right)+\frac{\operatorname{ctg}^{2} x-1}{2 \operatorname{ctg} x}>0 \Leftrightarrow$
$\Leftrightarrow \operatorname{ctg}\left(x+\frac{\pi}{3}\right)+\operato... | x\in(-\frac{\pi}{3}+\pin;-\frac{\pi}{9}+\pin)\cup(\pin;\frac{2\pi}{9}+\pin)\cup(\frac{\pi}{2}+\pin;\frac{5\pi}{9}+\pin),n\in\mathcal{Z} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,030 |
$9.286 \sin ^{3} x \cdot \sin \left(\frac{\pi}{2}-3 x\right)+\cos ^{3} x \cdot \cos \left(\frac{\pi}{2}-3 x\right)>\frac{3 \sqrt{3}}{8}$. | ## Solution.
$$
\begin{aligned}
& \sin ^{3} x \cdot \cos 3 x+\cos ^{3} x \cdot \sin 3 x>\frac{3 \sqrt{3}}{8} \Leftrightarrow \\
& \Leftrightarrow 2 \sin ^{2} x \cdot \sin x \cdot \cos 3 x+2 \cos ^{2} x \cdot \cos x \cdot \sin 3 x \geq \frac{3 \sqrt{3}}{4} \Leftrightarrow \\
& \Leftrightarrow(1-\cos 2 x) \sin x \cdot \... | x\in(\frac{\pi}{12}+\frac{\pin}{2};\frac{\pi}{6}+\frac{\pin}{2}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,031 |
$9.2872 \sin ^{2} x-\sin x+\sin 3 x<1$.
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$9.2872 \sin ^{2} x-\sin x+\sin 3 x<1$. | Solution.
$$
\begin{aligned}
& 2 \sin ^{2} x-\sin x+\sin 3 x-4 \sin ^{3} x=0 \Leftrightarrow \\
& \Leftrightarrow\left(2 \sin ^{2} x-1\right)(2 \sin x-1)>0 \Leftrightarrow\left(\sin x-\frac{1}{2}\right)\left(\sin x-\frac{\sqrt{2}}{2}\right)\left(\sin x+\frac{\sqrt{2}}{2}\right)>0 \Leftrightarrow \\
& \Leftrightarrow\l... | x\in(\frac{\pi}{4}+2\pin;\frac{3\pi}{4}+2\pin)\cup(-\frac{\pi}{4}+2\pin;\frac{\pi}{6}+2\pin)\cup(\frac{5\pi}{6}+2\pin;\frac{5\pi}{4}+2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,032 |
$9.288 \operatorname{ctg} x-\operatorname{tg} x-2 \operatorname{tg} 2 x-4 \operatorname{tg} 4 x>8 \sqrt{3}$. | Solution. $\left(\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}\right)-2 \operatorname{tg} 2 x-4 \operatorname{tg} 4 x>8 \sqrt{3} \Leftrightarrow$
$\Leftrightarrow \frac{\cos 2 x}{2 \sin x \cdot \cos x}-\operatorname{tg} 2 x-2 \operatorname{tg} 4 x>4 \sqrt{3} \Leftrightarrow(\operatorname{ctg} 2 x-\operatorname{tg} 2 x)-... | x\in(\frac{\pin}{8};\frac{\pi}{48}(6n+1)) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,033 |
$9.2894 \sin x \cdot \sin 2 x \cdot \sin 3 x>\sin 4 x$. | ## Solution.
$2 \sin x \cdot \sin 2 x \cdot \sin 3 x>\sin 2 x \cdot \cos 2 x \Leftrightarrow \sin 2 x(2 \sin x \cdot \sin 3 x-\cos 2 x)>0 \Leftrightarrow$ $\Leftrightarrow \sin 2 x(\cos 2 x-\cos 4 x-\cos 2 x)>0 \Leftrightarrow-\sin 2 x \cdot \cos 4 x>0 \Leftrightarrow$
$\Leftrightarrow\left(2 \sin ^{2} 2 x-1\right) \s... | x\in(-\frac{\pi}{8}+\pin;\pin)\cup(\frac{\pi}{8}+\pin;\frac{3\pi}{8}+\pin)\cup(\frac{\pi}{2}+\pin;\frac{5\pi}{8}+\pin) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,034 |
9.291 Prove the inequality: $\frac{1}{8}<\cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 70^{\circ}<\frac{1}{4}$. | ## Solution.
$\frac{1}{2}<4 \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \sin 20^{\circ}<1 \Leftrightarrow \frac{1}{2}<2 \sin 40^{\circ} \cdot \cos 40^{\circ}<1 \Leftrightarrow$
$\Leftrightarrow \frac{1}{2}<\sin 80^{\circ}<1 \Leftrightarrow \sin 30^{\circ}<\sin 80^{\circ}<\sin 90^{\circ}$.
The last statement is obvio... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,036 |
9.292 Solve the inequality: $\frac{\cos ^{2} 2 x}{\cos ^{2} x} \geq 3 \operatorname{tg} x$. | ## Solution.
$\frac{\cos ^{2} 2 x}{\cos ^{2} x} \geq \frac{3 \sin x}{\cos x} \Leftrightarrow\left\{\begin{array}{l}2 \cos ^{2} 2 x \geq 3 \sin 2 x, \\ x \neq \frac{\pi}{2}+\pi k\end{array} \Leftrightarrow\left\{\begin{array}{l}2 \sin ^{2} 2 x+3 \sin 2 x-2 \leq 0, \\ x \neq \frac{\pi}{2}+\pi k\end{array} \Leftrightarro... | x\in[-\frac{7\pi}{12}+\pin;-\frac{\pi}{2}+\pin)\cup(-\frac{\pi}{2}+\pin;\frac{\pi}{12}+\pin] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,037 |
9.293 Prove: $\boldsymbol{\operatorname { t g }} 40^{\circ}+\boldsymbol{\operatorname { t g }} 45^{\circ}+\boldsymbol{t g} 50^{\circ}>3$. | Solution. $\operatorname{tg} 40^{\circ}+1+\operatorname{tg} 50^{\circ}>3 \Leftrightarrow \operatorname{tg} 40^{\circ}+\operatorname{ctg} 40^{\circ}-2>0 \Leftrightarrow \frac{\left(\operatorname{tg} 40^{\circ}-1\right)^{2}}{\operatorname{tg} 40^{\circ}}>0$. The last inequality is obvious. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,038 |
9.294 Prove that for $360^{\circ} k-45^{\circ}<\alpha<360^{\circ} k+45^{\circ}, k \in Z$, the inequality $\operatorname{ctg}\left(45^{\circ}-\alpha\right)+\operatorname{ctg} 45^{\circ}+\operatorname{ctg}\left(45^{\circ}+\alpha\right) \geq 3$ holds. | Solution. $\operatorname{tg}\left(90^{\circ}-\left(45^{\circ}-\alpha\right)\right)+1+\frac{1}{\operatorname{tg}\left(45^{\circ}+\alpha\right)} \geq 3 \Leftrightarrow \frac{\left(\operatorname{tg}\left(45^{\circ}+\alpha\right)-1\right)^{2}}{\operatorname{tg}\left(45^{\circ}+\alpha\right)} \geq 0$.
By the condition, $36... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,039 |
$9.2953 \cos ^{2} x \cdot \sin x-\sin ^{3} x<\frac{1}{2}$. | Solution. $\sin x \cdot\left(3 \cos ^{2} x-\sin ^{2} x\right)<\frac{1}{2} \Leftrightarrow \sin x \cdot\left(3-4 \sin ^{2} x\right)<\frac{1}{2} \Leftrightarrow \sin 3 x<\frac{1}{2} \Leftrightarrow$ $\Leftrightarrow-\frac{7 \pi}{6}+2 \pi n<3 x<\frac{\pi}{6}+2 \pi n \Leftrightarrow-\frac{7 \pi}{18}+\frac{2 \pi n}{3}<x<\fr... | x\in(\frac{\pi}{18}(12n-7);\frac{\pi}{18}(12n+1)) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,040 |
$9.296 \frac{\cos x+2 \cos ^{2} x+\cos 3 x}{\cos x+2 \cos ^{2} x-1}>1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$9.296 \frac{\cos x+2 \cos ^{2} x+\cos 3 x}{\cos x+2 \cos ^{2} x-1}>1$. | Solution. $\frac{(\cos x+\cos 3 x)+2 \cos ^{2} x}{\cos x+\cos 2 x}>1 \Leftrightarrow \frac{\cos x \cdot(\cos 2 x+\cos x)}{\cos 2 x+\cos x}>\frac{1}{2} \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}\cos x>\frac{1}{2}, \\ 2 \cos ^{2} x+\cos x-1 \neq 0\end{array} \Leftrightarrow \cos x>\frac{1}{2} \Leftrightarro... | x\neq\frac{\pi}{2}(4n+1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,041 |
9.299 Prove the inequality: $\frac{1}{8}<\sin 20^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}<\frac{1}{4}$. | ## Solution.
$\frac{1}{2}<4 \sin 20^{\circ} \cdot \sin 50^{\circ} \cdot \cos 20^{\circ}<1 \Leftrightarrow \frac{1}{2}<2 \sin 40^{\circ} \cdot \cos 40^{\circ}<1 \Leftrightarrow$
$\Leftrightarrow \frac{1}{2}<\sin 80^{\circ}<1 \Leftrightarrow \sin 30^{\circ}<\sin 80^{\circ}<1$.
The last statement is obvious. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 52,043 |
9.300 Solve: $\log _{x^{2}-3} 729>3$. | ## Solution.
$$
\begin{aligned}
& \log _{x^{2}-3} 9^{3}>3 \Leftrightarrow \log _{x^{2}-3} 9>\log _{x^{2}-3}\left(x^{2}-3\right) \Leftrightarrow \\
& \Leftrightarrow\left[\begin{array} { l }
{ \{ \begin{array} { l }
{ 0 < x^2 - 3 < 1 , } \\
{ x^2 - 3 > 1 , } \\
{ 0 < 9 }
\end{array} \Leftrightarrow \left[\begin{array... | x\in(-\sqrt{12};-2)\cup(2;\sqrt{12}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,044 |
### 9.301 Solve the inequality: $\frac{\log _{a}\left(35-x^{3}\right)}{\log _{a}(5-x)}>3$. | Solution. $\log _{5-x}\left(35-x^{3}\right)>3 \Leftrightarrow \log _{5-x}\left(35-x^{3}\right)>\log _{5-x}(5-x)^{3} \Leftrightarrow$ $\Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}5-x>1, \\ 35-x^{3}>(5-x)^{3}, \\ 00\end{array}\right.\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}x... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,045 |
### 9.302 Find the domain of the function
$y=\sqrt{\log _{1 / 4}\left(\frac{x}{x+1}\right)^{2}-1}$. | ## Solution.
$\log _{1 / 4}\left(\frac{x}{x+1}\right)^{2} \geq \log _{1 / 4} \frac{1}{4} \Leftrightarrow\left\{\begin{array}{l}\frac{x^{2}}{(x+1)^{2}} \leq \frac{1}{4}, \\ x \neq 0 .\end{array} \Leftrightarrow\left\{\begin{array}{l}3 x^{2}-2 x-1 \leq 0, \\ x \neq 0, \\ x \neq-1\end{array} \Leftrightarrow\left\{\begin{... | x\in[-\frac{1}{3};0)\cup(0;1] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 52,046 |
### 9.303 Find integer values of $x$ that satisfy the inequality
$$
\log _{0.3}(\sqrt{x+5}-x+1)>0
$$ | Solution. The inequality is equivalent to the system:
$\left\{\begin{array}{l}\sqrt{x+5}-x+10, \\ x-\text { integer }\end{array} \Leftrightarrow\left\{\begin{array}{l}\sqrt{x+5}x-1, \\ x-\text { integer. }\end{array}\right.\right.$
From this, $x>0$ and, consequently, $x \geq 1$ ( $x$ - integer). Therefore,
$\left\{\... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,047 |
9.304 Find $x$, for which the inequality $y^{2}-\left(5^{x}-1\right)(y-1)>0$ holds for all $y$. | ## Solution.
$y^{2}-\left(5^{x}-1\right) y+5^{x}-1>0$. This inequality holds for all $y$ if and only if $D<0$, where
$D=\left(5^{x}-1\right)^{2}-4 \cdot 5^{x}+4=\left(5^{x}-1\right)\left(5^{x}-5\right)<0 \Leftrightarrow 1<5^{x}<5 \Leftrightarrow 0<x<1$.
Answer: $x \in(0 ; 1)$. | x\in(0;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,048 |
### 9.305 Find $a$, for which the inequality
$x^{2}-2^{a+2} \cdot x-2^{a+3}+12>0$ is true for any $x$. | Solution. The requirement of the problem is equivalent to the condition: the discriminant $D$ is less than 0, where $D=\left(2^{a+2}\right)^{2}-4\left(12-2^{a+3}\right)=16\left(2^{2 a}+2 \cdot 2^{a}-3\right)$. $D<0 \Leftrightarrow 2^{2 a}+2 \cdot 2^{a}-3<0 \Leftrightarrow\left(2^{a}+3\right)\left(2^{a}-1\right)<0 \Left... | \in(-\infty;0) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 52,049 |
10.361 In triangle $A B C$, the measure of angle $A$ is twice the measure of angle $B$, and the lengths of the sides opposite these angles are 12 cm and 8 cm, respectively. Find the length of the third side of the triangle. | Solution.
A
Let $A K$ be the bisector of $\angle A$. Then
$\angle K A C = \angle A B C \Rightarrow \triangle A K C \sim \triangle A B C \Rightarrow$
$\Rightarrow \frac{A C}{B C} = \frac{K C}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,050 |
10.362 The hypotenuse of a right triangle is equal to $m$, the radius of the inscribed circle is $r$. Determine the legs. Under what relation between $r$ and $\boldsymbol{m}$ does the problem have a solution? | Solution. Let $x, y$ be the legs of the given triangle. Then, by Theorem 1 and the Pythagorean Theorem, we have:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ r = \frac { x + y - m } { 2 } } \\
{ m ^ { 2 } = x ^ { 2 } + y ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 2 r + m = x + y , } \\
{... | \frac{2r+\\sqrt{^{2}-4r^{2}-4r}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,051 |
10.363 In an isosceles triangle with a base of 12 cm, a circle is inscribed, and three tangents are drawn to it such that they cut off three small triangles from the given triangle. The sum of the perimeters of the small triangles is 48 cm. Find the lateral side of the given triangle. | Solution.
Answer: $18 \mathrm{~cm}$. | 18\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,052 |
10.364 In an isosceles triangle, a circle is inscribed. The points of tangency divide each lateral side into segments of length $\boldsymbol{m}$ and $\boldsymbol{n}$, measured from the vertex. Three tangents are drawn to the circle, parallel to each of the triangle's sides. Find the lengths of the segments of the tange... | ## Solution.
Let the tangents $M N, K L$ to the circle inscribed in the given $\triangle A B C$ with base $A C$ pass through points $E$ and $F$ of the circle; $P, Q, R$ are the points of ta... | \frac{2n}{+2n};\frac{n(+n)}{+2n} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,053 |
10.365 Determine the acute angles of a right triangle if the ratio of the radii of the circumscribed and inscribed circles is $\sqrt{3}+1$. | Solution. Let $a, b$ be the legs of the given triangle, $c$ be its hypotenuse, and $R$ and $r$ be the radii of the circumscribed and inscribed circles. Then $R=\frac{c}{2}, r=\frac{a+b-c}{2}$ (Theorem 1) $\Rightarrow$ $\Rightarrow \frac{r}{R}=\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}=\frac{a+b-c}{c}==\frac{a}{c}+\frac{... | 30,60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,054 |
10.366 Two circles touch each other externally at point $A$. Find the radii of the circles if the chords connecting point $A$ with the points of tangency of one of their common external tangents are 6 and 8 cm. | ## Solution.
Let $BC$ be the common tangent to the two given circles $O_1$ and $O_2$ ($B$ and $C$ are the points of tangency), $AB = 8$, $AC = 6$. Draw the tangent at point $A$ until it in... | \frac{15}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,055 |
10.367 Express the side of a regular decagon in terms of the radius $R$ of the circumscribed circle. | ## Solution.
Let $AB$ be a side of the given regular decagon with center $O$, then $AO = BO = R$, $\angle A = \angle B = 72^{\circ}$. Draw the angle bisector $AK$.
Then $\angle AOK = \angle... | \frac{\sqrt{5}-1}{2}R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,056 |
10.368 Calculate the length of the angle bisector of angle $A$ in triangle $A B C$ with side lengths $a=18$ cm, $b=15$ cm, $c=12$ cm. | Solution. Let $A K$ be the bisector of $\angle A$ of the given $\triangle A B C$. By Theorem 4, $\frac{C K}{B K}=\frac{15}{12} \Rightarrow \frac{18-B K}{B K}=\frac{15}{12} \Rightarrow B K=8, C K=10$.
Applying the Law of Cosines to $\triangle A B K$ and $\triangle A C K$, we have:
$\left\{\begin{array}{l}8^{2}=12^{2}+... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,057 |
10.369 In a triangle with a perimeter of 20 cm, a circle is inscribed. The segment of the tangent, drawn parallel to the base and enclosed between the sides of the triangle, measures 2.4 cm. Find the base of the triangle. | ## Solution.
Answer: 4 cm or 6 cm. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,058 |
### 10.370 The larger of the parallel sides of a trapezoid is equal to $a$, the smaller is equal to $b$, the non-parallel sides are equal to $c$ and $d$. Find the area of the trapezoid. | ## Solution.
$$
=\frac{2 b \cdot S_{K C D}}{a-b} \Rightarrow S_{A B C D}=\begin{aligned}
& \text { Let } A B C D-\text { the given trapezoid, } \\
& \begin{array}{l}
A D=a ; B C=b, A B=c, C D=d, h-\text { height } \\
\text { of the trapezoid. Draw } C K \| A B . \\
\text { Then } K D=a-b, C K=c
\end{array} \\
& h=\fra... | \frac{+b}{4(-b)}\cdot\sqrt{(++-b)(+-b-)(+-b-)(b++-)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,059 |
10.371 Point $C_{1}$ is the foot of the altitude $C C_{1}$ of triangle $A B C$. Find the relationship between angles $A$ and $B$ if $C C_{1}^{2}=C_{1} A \cdot C_{1} B$. | Solution. By the given condition, $\frac{C C_{1}}{A C_{1}}=\frac{B C_{1}}{C C_{1}} \Rightarrow \triangle A C C_{1} \sim \triangle B C C_{1} \Rightarrow \angle C A C_{1}=\angle B C C_{1}$.
If now $\angle A=90^{\circ}$, then $180^{\circ}-\angle A=\angle C A C_{1}=\angle C_{1} C B=90^{\circ}-\angle B \Rightarrow$
$\Righ... | \angleA+\angleB=90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,060 |
10.372 The bisector of angle $A$ of triangle $A B C$ intersects the circumscribed circle around it at point $D$. Find the length of the chord $D C$, if the center of the circle inscribed in the given triangle is at a distance $\boldsymbol{n}$ from point $D$. | Solution.
D
Let $O$ be the center of the circle inscribed in the given $\triangle ABC$. $\angle BCD = \angle BAD$ (Theorem 6) $\Rightarrow$ $\Rightarrow \angle BCD + \angle OCB = \angle CAD ... | n | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,061 |
10.373 In a triangle with sides 6, 10, and 12 cm, a circle is inscribed. A tangent is drawn to the circle such that it intersects the two larger sides. Find the perimeter of the cut-off triangle. | ## Solution.
Let $M N$ be the tangent to the circle inscribed in the given $\triangle A B C, A B=10, B C=12, A C=6$.
Then $B M+B N+M N=$ $=(10-A M)+(12-C N)+(A M+C N-6)=16$ (by theorem $2, ... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,062 |
10.374 The bases of an isosceles trapezoid are 4 and 8 cm, and its area is 21 cm². Which side does the bisector of the angle at the larger base intersect: the smaller base or the lateral side of the trapezoid? | ## Solution.
Let $C K$ be the height of the given isosceles trapezoid $A B C D, B C=4, A D=8$. Then
$$
\begin{aligned}
& K D=\frac{A D-B C}{2}=2, C K=\frac{2 S_{A B C D}}{4+8}=\frac{7}{2} ... | thelateralside | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,063 |
10.375 An equilateral triangle $ABC$, inscribed in a circle of radius $R$, is rotated around the center of the circle by $90^{\circ}$ to the position $A_{1} B_{1} C_{1}$. Calculate the area of the hexagon $A A_{1} B B_{1} C C_{1}$. | Solution. Let $O$ be the center of the circumcircle of the given equilateral $\triangle A B C$. Then the hexagon $A A_{1} B B_{1} C C_{1}$ will be composed of three equal triangles $\triangle O A A_{1}, \triangle O B B_{1}, \triangle O C C_{1}$ and three equal triangles
$\triangle O A_{1} B, \triangle O B_{1} C, \tria... | \frac{9R^{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,064 |
10.377 The bisector of a triangle's angle divides the opposite side into segments of length 4 and 2 cm, and the height drawn to the same side is $\sqrt{15}$ cm. What are the lengths of the sides of the triangle, given that they are expressed as integers? | Solution. Let the bisector of angle $A$ divide side $B C$ of the given $\triangle A B C$ into segments 2 and 4. Suppose $A Bh=\sqrt{15}(h$ - the height of $\triangle A B C) \Rightarrow \sqrt{15}<n<6 \Rightarrow 4 \leq n \leq 5$.
If $n=5$, then by Heron's formula $S_{A B C}=\sqrt{\frac{21}{2} \cdot \frac{11}{2} \cdot \... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,065 |
10.378 Two circles touch each other externally. Four points of tangency of their external common tangents $A, B, C, D$ are connected sequentially. Show that a circle can be inscribed in the quadrilateral $\boldsymbol{A} \boldsymbol{B} C D$, and find its radius if the radii of the given circles are $\boldsymbol{R}$ and ... | Solution.
Let $K$ be the point of tangency of the two given circles with centers $Q$ and $P$. Draw the common tangent through $K$ until it intersects the external tangents $A B$ and $C D$ a... | \frac{2Rr}{R+r} | Geometry | proof | Yes | Yes | olympiads | false | 52,066 |
10.381 The height of the triangle, equal to 2 cm, divides the angle of the triangle in the ratio $2: 1$, and the base of the triangle - into parts, the smaller of which is equal to 1 cm. Determine the area of the triangle. | ## Solution.
Answer: $\frac{11}{3} \mathrm{~cm}^{2}$. | \frac{11}{3}\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,069 |
10.382 Given two concentric circles. Prove that the sum of the squares of the distances from a point on one circle to the ends of a diameter of the other circle does not depend on the chosen point or the chosen diameter. | Solution.
Let $AB$ be the diameter of one of the given concentric circles with radius $R_{1}$,
$C$ be an arbitrary point on the other circle with radius $R_{2}$, and $O$ be their common ce... | 2R_{1}^{2}+2R_{2}^{2} | Geometry | proof | Yes | Yes | olympiads | false | 52,070 |
### 10.383 In triangle $ABC$, medians $AL$ and $BM$ are drawn, intersecting at point $K$. Vertex $C$ lies on the circle passing through points $K, L, M$. Show that the median $CN$ forms the same angles with sides $AC$ and $BC$ as medians $BM$ and $AL$ form with side $AB$. | ## Solution.
$$
\angle K C M=\angle K L M \text { (Theorem 6), } \angle K L M=\angle N A K
$$
(since $L M \| A B$ ) $\Rightarrow \angle K C M=\angle N A K$, which is what we needed to prov... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,071 |
### 10.384 In triangle $A B C$, the bisectors $A D$ and $C E$ intersect at
point $F$. Points $B, D, E, F$ lie on the same circle. Show that angle $B$ is $60^{\circ}$. | ## Solution.
$$
\begin{aligned}
& \angle E F D=180^{\circ}-(\angle F A C+\angle F C A)= \\
& =180^{\circ}-\frac{1}{2}(\angle A+\angle C) \Rightarrow 2 \angle B=\angle A+\angle C \Rightarro... | 60 | Geometry | proof | Yes | Yes | olympiads | false | 52,072 |
10.385 The area of the triangle is S. Each side of the triangle is divided into three parts in the ratio $m: n: m$. Determine the area of the hexagon whose vertices are the points of division. | Solution.
Let points $K, M, N, L, Q, P$ divide the sides of the given $\triangle A B C$ in the ratio $m: n: m$.
$$
\begin{gathered}
\text { Then } \frac{B M}{A B}=\frac{m}{2 m+n}=\frac{B ... | S(1-\frac{3^{2}}{(2+n)^{2}}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,073 |
### 10.386 The distances from the center of the circle inscribed in a right-angled triangle to the vertices of its acute angles are $\sqrt{5}$ and $\sqrt{10}$. Find the legs. | ## Solution.
$B M=\sqrt{O B^{2}-1}=3, C N=\sqrt{O C^{2}-1}=2 \Rightarrow A B=r+B M=4, A C=r+C N=3$ (AMON - square).
Answer: 3 and 4. | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,074 |
10.387 In triangle $A B C$, each altitude $h_{c}$ and $h_{b}$ is not less than the side to which it is dropped. Find the angles of the triangle. | Solution. Let $\alpha=\angle B A C, c=A B, b=A C$. Then
$S_{A B C}=\frac{b \cdot c \cdot \sin \alpha}{2}=\frac{b \cdot h_{b}}{2}=\frac{c \cdot h_{c}}{2} \Rightarrow b \cdot \sin \alpha=h_{c}, c \cdot \sin \alpha=h_{b} \Rightarrow$
$\Rightarrow \sin \alpha \cdot(b+c)=h_{b}+h_{c}$. By the condition, $h_{b} \geq b, h_{c... | 90,45,45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,075 |
10.389 In an isosceles triangle $A B C \quad A B=B C=25$ cm and $A C=14$ cm. Calculate the radius of the circle that touches $B C$ at point $D$, the foot of the altitude $A D$, and passes through the midpoint of $A C$. | ## Solution.
Let $O$ be the center of the circle passing through the midpoint $K$ of the base $AC$ and touching the side $BC$ at point $D$. Clearly, $BK = \sqrt{AB^2 - AK^2} = 24$.
$AC \cd... | \frac{175}{48} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,076 |
10.391 On the segment $A C$ there is a point $B$, and $A B=14 \text{~cm}, B C=28 \text{~cm}$. On the segments $A B, B C$ and $A C$ as diameters, semicircles are constructed in the same half-plane relative to the boundary $\boldsymbol{A} \boldsymbol{B}$. Find the radius of the circle that is tangent to all three semicir... | ## Solution.
Let $O_{1}, O_{2}, O_{3}$ be the centers of semicircles with radii $r_{1}, r_{2}, r_{3}$ and diameters $A B, B C$, and $A C$ respectively; $P$ be the center of the circle with... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,077 |
10.392 In a circle of radius $R$, an equilateral triangle and a square are inscribed, sharing a common vertex. Calculate the area of the common part of the triangle and the square. | ## Solution.
Let the square $ABCD$ and the equilateral $\triangle AEF$ be inscribed in a given circle of radius $R$. Let $M$, $N$, $K$, $L$ be the points of intersection of the sides of the... | \frac{R^2(8\sqrt{3}-9)}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,078 |
### 10.393 An isosceles trapezoid is circumscribed around a circle with radius $R=1$ cm, and the area of the trapezoid is $5 \mathrm{~cm}^{2}$. Find the area of the quadrilateral whose vertices are the points of tangency of the circle and the trapezoid. | ## Solution.
A
Let $O$ be the center of the circle circumscribed around the given isosceles trapezoid $ABCD$, and let $K, L, M, N$ be the points of tangency with the circle. $\angle BAO = ... | 1.6^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,079 |
10.394 Inside triangle $A B C$, a random point is taken, and three lines are drawn through it, parallel to the sides of the triangle. These lines divide triangle $A B C$ into six parts, three of which are triangles. The areas of these triangles are $S_{1}, S_{2}$, and $S_{3}$. Prove that the area of triangle $A B C$ is... | Solution.
 | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,080 |
### 10.395 The centers of four circles are located at the vertices of a square
with side $a$. The radii of these circles are $a$. Determine the area of their common part. | ## Solution.
$C$
B
A
$$
S_{A M N}=\frac{a^{2}}{2} \sin 30^{\circ}=\frac{a^{2}}{4} \Rightarrow S_{\dot{M} N N}=S_{A M P N}-S_{A M N}=\frac{a^{2}}{4}\left(\frac{\pi}{3}-1\right)
$$
The com... | ^{2}\cdot\frac{\pi+3-3\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,081 |
10.396 Three equal circles are inscribed in a circle of radius $R$, touching the outer circle and each other. Calculate the area of the figure bounded by these three circles. | Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of three mutually tangent circles of radius $r$, inscribed in a given circle of radius $R$ with center $O_{4}$.
Obviously, $O_{4}$ coincides with the center of the equilateral $\triangle O_{1} O_{2} O_{3}$ with side $2 r$, and $O_{4} O_{1}=R-r$ (the line $O_{4} O_{1}$... | \frac{1}{2}R^{2}(7-4\sqrt{3})(6\sqrt{3}-3\pi) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,082 |
10.397 Six equal circles are inscribed in a circle of radius $R$, each touching the given circle and two adjacent ones. Calculate the area of the figure bounded by these six circles. | Solution. Let $O_{1}, o_{2}, o_{3}, O_{4}, o_{5}, o_{6}$ be the centers of six circles of radius $r$, each touching two adjacent circles and inscribed in a given circle of radius $R$ with center $P$. Clearly, $P$ coincides with the center of the regular hexagon $T$ with vertices $O_{i}, i=1,2, \ldots 6$, and side lengt... | \frac{2}{9}R^{2}(3\sqrt{3}-\pi) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,083 |
10.398 Four equal circles are inscribed in a circle of radius $R$, each touching the given circle and two adjacent ones. Calculate the area of the figure bounded by these four circles. | Solution: Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the centers of four circles of radius $r$, each touching two adjacent ones, and inscribed in a circle of radius $R$ with center $P$. Clearly, $P$ coincides with the center of the square $O_{1} O_{2} O_{3} O_{4}$ with side $2 r$. Moreover, $P O_{1}=R-r$ (the line $P O_{1}$ p... | R^{2}(4-\pi)(3-2\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,084 |
10.399 The side of a regular triangle is equal to a. A circle with a radius of a/3 is described from its center. Determine the area of the part of the triangle that lies outside the circle. | ## Solution.
Let $O$ be the center of the given equilateral $\triangle ABC$, $M$ and $N$ be the points of intersection of the circle with center $O$ and radius $a / 3$ with sides $AB$ and ... | \frac{^2(3\sqrt{3}-\pi)}{18} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,085 |
10.400 Calculate the area of a triangle given two sides $\boldsymbol{a}$ and $\boldsymbol{b}$ and the bisector $l$ of the angle between them. | Solution.
Let $B D$ be the bisector of the given $\triangle A B C, a=A B, b=B C, l=B D, \angle B=2 \alpha$. By the cosine rule for $\triangle A B D$ and $\triangle D B C$, we have:
$$
\lef... | \frac{(+b)}{4}\sqrt{4^{2}b^{2}-^{2}(+b)^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,086 |
10.401 Find the radius of the circle if the area of the circle is $Q$ square units greater than the area of the inscribed regular dodecagon. | Solution. A regular 12-sided polygon inscribed in a given circle of radius $R$ consists of 12 identical isosceles triangles with legs $R$ and an angle of $30^{\circ}$ at the vertex. Therefore, $Q=\pi R^{2}-\frac{12 R^{2} \sin 30^{\circ}}{2} \Rightarrow R=\sqrt{\frac{Q}{\pi-3}}$.
Answer: $\sqrt{\frac{Q}{\pi-3}}$. | \sqrt{\frac{Q}{\pi-3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,087 |
10.402 A circle of radius $R$ with center at point $O$ is divided into six equal parts by points $A, B, C, D, E, F$. Determine the area of the figure $C O E$, bounded by the arc $O C$ with center at point $B$, the arc $O E$ with center at point $F$, and the arc $C E$ with center at point $A$. | ## Solution.
The area of the desired figure $\Phi$, bounded by
, the altitude $CD$ is drawn. The radii of the circles inscribed in triangles $ACD$ and $BCD$ are 0.6 and 0.8 cm, respectively. Find the radius of the circle inscribed in triangle $\boldsymbol{A} \boldsymbol{\text { B }}$. | Solution. Let $x, r_{1}, r_{2}$ be the radii of the circles inscribed in $\triangle A B C$, $\triangle A C D, \triangle B C D$ respectively. $\triangle A C D \sim \triangle B C D \Rightarrow \frac{A C}{B C}=\frac{r_{1}}{r_{2}}=\frac{0.6}{0.8}$ (the radii of the circles inscribed in similar triangles are proportional to... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,089 |
10.404 The area of triangle $ABC$ is $S_{1}$; the area of triangle $AOB$, where $O$ is the orthocenter, is $S_{2}$. On the line $CO$, a point $K$ is taken such that triangle $ABK$ is a right triangle. Prove that the area of triangle $ABK$ is the geometric mean between $S_{1}$ and $S_{2}$. | Solution.
Let $CD$ be the height of the given $\triangle ABC$.
$$
\begin{aligned}
& \triangle BCD \sim \triangle ADO \quad\left(\angle BCD = \angle BAO = 90^\circ - \angle B\right) \Righta... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,090 |
10.405 A circle is inscribed in an equilateral triangle with side $a$. A tangent is drawn to the circle such that the segment of it inside the triangle is equal to $b$. Find the area of the triangle cut off by this tangent from the given one. | ## Solution.
Let $M N$ be the tangent to the circle inscribed in the given equilateral $\triangle A B C, M N=b$. Let
$$
\begin{gathered}
x=B M, y=B N . A M+C N=M N+A C=a+b \\
(\text { The... | \frac{(-2b)\sqrt{3}}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,091 |
10.407 The sides of triangle $A B C$ are divided by points $M, N$ and $P$ such that $A M: M B=B N: N C=C P: P A=1: 4$. Find the ratio of the area of the triangle bounded by the lines $A N, B P$ and $C M$ to the area of triangle $\boldsymbol{A} \boldsymbol{B} C$. | ## Solution.
Let $x = S_{A M K}$, then $S_{B M K} = 4x$ (Theorem 10).
\[
\begin{aligned}
& S_{A B K} = 5x \\
& S_{A N C} = 4 S_{A N B}, S_{K N C} = 4 S_{K N B} (\text{Theorem } 10) \Right... | \frac{3}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,093 |
10.408 An isosceles trapezoid is circumscribed around a circle with a radius of 5 cm. The distance between the points of tangency of its lateral sides is 8 cm. Find the area of the trapezoid. | Solution.
Let $O$ be the center of the circle with radius 5, around which an isosceles trapezoid $ABCD$ is circumscribed, and $N, L, M, K$ are the points of tangency with the circle.
$$
\begin{gathered}
B M=B N=C N=C L(\text { theorem } 3) \Rightarrow \\
\Rightarrow M L \| B C(\text { Thales' theorem }) \Rightarrow \... | 125^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,094 |
10.409 A triangle with sides 13, 14, and 15 is divided into three equal areas by lines perpendicular to the longest side. Find the distances from these lines to the nearest vertices of the triangle located on the longest side. | ## Solution.
Let in $\triangle ABC$, $AC=15$, $BC=14$, $AB=13$, $MN \perp AC$, $KL \perp AC$,
$S_{AMN}=S_{B K L M N}=S_{K C L}$. By the cosine rule for $\triangle ABC$, $\cos \angle A=\frac... | \sqrt{33},\sqrt{42} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,095 |
10.410 A circle is inscribed in a trapezoid, where the smaller base is equal to $a$. One of the lateral sides of the trapezoid is divided by the point of tangency into segments of length $\boldsymbol{m}$ and $n$, measured from the larger base. Determine the area of the trapezoid. | Solution.
Let $O$ be the center of the circle with radius $r$ inscribed in the given trapezoid $ABCD$. $M, N, K, L$ are the points of tangency of the trapezoid with the circle. $BC = a, AM ... | \frac{\sqrt{\cdotn}\cdot(n+)}{n} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,096 |
10.411 Two regular triangles with area $S$ are given, where the second triangle is obtained by rotating the first triangle around its center by an angle of $30^{\circ}$. Calculate the area of the intersection of these triangles. | ## Solution.
Let the regular triangles $\triangle ABC$ and $\triangle A_1B_1C_1$ have a common center $O$, and $\triangle A_1B_1C_1$ is rotated around $O$ relative to $\triangle ABC$ by $30... | S(\sqrt{3}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,097 |
10.412 The area of a right triangle is equal to $2 r^{2} / 3$, where $r$ is the radius of the circle that touches one leg and the extension of the other leg and the hypotenuse. Find the sides of the triangle. | ## Solution.
B
A
K
0
Let the circle with center $O$ and radius $r$ touch the leg $A C$ and the extensions of the leg $A B$ and the hypotenuse $B C$ of the given $\triangle A B C$ at points... | r,\frac{4}{3}r,\frac{5}{3}r | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,098 |
10.413 The diagonals of a trapezoid divide it into four triangles. Prove that if the areas of two of them, adjacent to the bases of the trapezoid, are equal to $p^{2}$ and $q^{2}$, then the area of the trapezoid is equal to $(p+q)^{2}$. | ## Solution.
Similarly, it is shown that $S_{C O D}=p \cdot q \Rightarrow S_{A B C D}=p^{2}+q^{2}+2 p q=$ $=(p+q)^{2}$, which is what we needed to prove. | (p+q)^2 | Geometry | proof | Yes | Yes | olympiads | false | 52,099 |
10.414 In quadrilateral $A B C D$, a line parallel to the other diagonal $A C$ is drawn through the midpoint of diagonal $B D$. This line intersects side $A D$ at point $E$. Prove that segment $C E$ divides quadrilateral $A B C D$ into equal areas. | ## Solution.
Let $M$ be the midpoint of diagonal $BD$, and $EM \parallel AC$. Extend $AD$ to point $K$ such that $KE = ED$. Then $S_{CED} = S_{CEK}$ (Theorem 10). $ME$ is the midline of $\t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 52,100 |
10.415 A line parallel to the bases of a given rectangular trapezoid divides it into two trapezoids, into each of which a circle can be inscribed. Find the bases of the original trapezoid, if its lateral sides are equal to $c$ and $d$, where $c<d$. | ## Solution.
Let in the given rectangular trapezoid $ABCD$, $c=AB$, $d=CD$, $MN \parallel BC$, and in $MBCN$ and $AMND$ inscribed circles with radii $r_1$ and $r_2$ can be inscribed. Extend... | \frac{(\sqrt{+}\\sqrt{-})^2}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,101 |
10.416 Determine the area of a triangle given its three heights $h_{1}$, $h_{2}, h_{3} \cdot$ | Solution. Let $a, b, c$ be the sides of the given triangle, to which the heights $h_{1}, h_{2}, h_{3}$ are drawn, respectively. Then $a=\frac{2 S}{h_{1}}, b=\frac{2 S}{h_{2}}, c=\frac{2 S}{h_{3}}$, where $S$ is the area of the given triangle. Substituting these expressions into Heron's formula, we have:
$$
S=\sqrt{\le... | \sqrt{(\frac{1}{h_{1}}+\frac{1}{h_{2}}+\frac{1}{h_{3}})(\frac{1}{h_{1}}+\frac{1}{h_{2}}-\frac{1}{h_{3}})(\frac{1}{h_{1}}+\frac{1}{h_{3}}-\frac{1}{h_{2}})(} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,102 |
### 10.417
In a right triangle $ABC\left(\angle C=90^{\circ}\right)$, a circle is inscribed, touching its sides at points $A_{1}, B_{1}, C_{1}$. Find the ratio of the area of triangle $ABC$ to the area of triangle $A_{1} B_{1} C_{1}$, if $AC=4 \text{ cm}, BC=3 \text{ cm}$. | ## Solution.
Let $O$ be the center of the circle with radius $r$ inscribed in the given $\triangle ABC$.
Then $AB=5, r=\frac{3+4-5}{2}=1$ (Theorem 1).
$S_{A_{1} B_{1} C_{1}}=S_{O A_{1} B_{1}}... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,103 |
10.419 A circle with its center on side $AB$ of triangle $ABC$ touches the other two sides of the triangle. Find the area of the circle if $a=13$ cm, $b=14$ cm, $c=15 \mathbf{cm}$, where $a, b$ and $c$ are the lengths of the sides of the triangle. | ## Solution.
Let $O$ be the center of the circle with radius $r$, touching
sides $A C$ and $B C$ of the given $\triangle A B C$ at points $K$ and $M$. By the cosine rule, $\cos \angle C=\fr... | \frac{3136}{81}\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,105 |
10.421 The height of the triangle, equal to 2 cm, divides the angle of the triangle in the ratio $2: 1$, and the base of the triangle into parts, the smaller of which is 1 cm. Find the area of this triangle. | ## Solution.
$A D=1.2 \cdot \angle A B D=\angle D B C$. Draw the bisector of angle $D B C$. Then $\angle A B D=\angle D B K=\angle K B C \Rightarrow$ $D K=A D=1$. Let $x=K C$. By Theorem 4, ... | \frac{11}{3}^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,107 |
10.422 In a right-angled triangle, the bisector of the right angle cuts off segments of length $a$ and $b$ on the hypotenuse. Find the area of the square whose side is this bisector. | ## Solution.
Let $AD$ be the bisector of the right angle of the given $\triangle ABC$, $BD=a$, $DC=b$. Drop perpendiculars $DM$ and $DN$ to the legs $AB$ and $AC$. Then $AMDN$ is a square ... | \frac{2^{2}b^{2}}{^{2}+b^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,108 |
10.424 From each vertex of the base of an equilateral triangle with side $a$, two rays are drawn, forming angles of $15^{\circ}$ and $30^{\circ}$ with this base. Find the area of the quadrilateral whose vertices are the points of intersection of the constructed rays. | ## Solution.
Let $O$ be the center of the given equilateral $\triangle ABC$. Then $AO$ and $CO$ are two given rays forming angles of $30^{\circ}$ with $AC$. Let the other two rays $AN$ and ... | \frac{^2(9-5\sqrt{3})}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,109 |
10.425 A chord $AB$ is drawn through a point $M$ located on the diameter of a circle with a radius of 4 cm, forming an angle of $30^{\circ}$ with the diameter. A chord $BC$ is drawn through point $B$, perpendicular to the given diameter. Find the area of triangle $ABC$, if $AM: MB=2: 3$. | ## Solution.
Let $M D \perp B C$. According to the problem, segment $M D$ lies on the diameter of the given circle. Therefore, $M D$ bisects chord $B C$ $\Rightarrow$ $\Rightarrow M B=M C \... | \frac{180\sqrt{3}}{19}\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,110 |
11.196 Two equal cubes with edge a have a common segment $E F$, the ends of which are the midpoints of two opposite edges, not belonging to the same face. One of the cubes is obtained by rotating the other around the line $\boldsymbol{E F}$ by $90^{\circ}$. Find the volume of the common part of these cubes. | ## Solution.
Let there be a cube $T_{1}=A B C D K L M N$ and a cube $T_{2}$, obtained by rotating $T_{1}$ around $E F$ by $90^{\circ}$. Let $P_{1}$ and $P_{2}$ be points on the edges $ND$ a... | \frac{^{3}(3\sqrt{2}-2)}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,111 |
11.197 Find the volume of the common part of two cubes, if one of them is obtained by rotating the other cube by $90^{\circ}$ around an axis passing through the midline of one of its faces. The edge of the cube is $a$. | ## Solution.
Let $EF$ be the midline of the face $ABCD$ of the first cube, from which the second cube is obtained by a $90^{\circ}$ rotation around $FE$ clockwise (when viewed from $F$ to $... | \frac{^{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 52,112 |
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