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 |
|---|---|---|---|---|---|---|---|---|---|
9.024. $\log _{3} \frac{3 x-5}{x+1} \leq 1$. | Solution.
The inequality is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ \frac { 3 x - 5 } { x + 1 } > 0 , } \\
{ \frac { 3 x - 5 } { x + 1 } \leq 3 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ ( x - \frac { 5 } { 3 } ) ( x + 1 ) > 0 , } \\
{ - \frac { 1 5 } { x + 1 } \leq ... | x\in(\frac{5}{3};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,070 |
9.026. $\log _{0.3}(3 x-8)>\log _{0.3}\left(x^{2}+4\right)$. | Solution.
This inequality is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ 3 x - 8 > 0 } \\
{ 3 x - 8 \frac{8}{3} \\
x^{2}-3 x+12>0
\end{array}\right.\right.
$$
Since $x^{2}-3 x+12>0$ for $x \in R$, the last system of inequalities is equivalent to the inequality $x>\frac{8}{3}$.
Answer: ... | x\in(\frac{8}{3};\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,071 |
9.027. $(x+1)(3-x)(x-2)^{2}>0$. | Solution.
We have
$$
(x+1)(x-3)(x-2)^{2}<0,\left\{\begin{array} { l }
{ ( x + 1 ) ( x - 3 ) < 0 , } \\
{ x - 2 \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
-1<x<3 \\
x \neq 2
\end{array}\right.\right.
$$
Answer: $\quad x \in(-1 ; 2) \cup(2 ; 3)$. | x\in(-1;2)\cup(2;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,072 |
9.030. $\frac{1}{x+2}<\frac{3}{x-3}$. | Solution.
We have
$\frac{1}{x+2}-\frac{3}{x-3}>0$.
Using the interval method, we get $x \in\left(-\frac{9}{2} ;-2\right) \cup(3 ; \infty)$.
$\xrightarrow[-\sqrt{+} \sqrt{-2} \sqrt{t}]{3} \sqrt{3}$
Answer: $x \in\left(-\frac{9}{2} ;-2\right) \cup(3 ; \infty)$. | x\in(-\frac{9}{2};-2)\cup(3;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,074 |
9.031. $\frac{3 x^{2}-10 x+3}{x^{2}-10 x+25}>0$. | Solution.
From the condition $\frac{3\left(x-\frac{1}{3}\right)(x-3)}{(x-5)^{2}}>0 \Leftrightarrow\left\{\begin{array}{l}\left(x-\frac{1}{3}\right)(x-3)>0, \\ x \neq 5 .\end{array}\right.$
Using the number line, we get $x \in\left(-\infty ; \frac{1}{3}\right) \cup(3 ; 5) \cup(5 ; \infty)$.
\cup(3;5)\cup(5;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,075 |
9.032. $\left|2 x^{2}-9 x+15\right| \geq 20$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.032. $\left|2 x^{2}-9 x+15\right| \geq 20$. | ## Solution.
We have $2 x^{2}-9 x+15>0$ for $x \in R$, therefore, the original inequality is equivalent to the inequality $2 x^{2}-9 x+15 \geq 20, 2 x^{2}-9 x-5 \geq 0$, $2(x-5)\left(x+\frac{1}{2}\right) \geq 0$. From this, $x \leq-\frac{1}{2}$ or $x \geq 5$.
Answer: $\quad x \in\left(-\infty ;-\frac{1}{2}\right] \cu... | x\in(-\infty;-\frac{1}{2}]\cup[5;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,076 |
9.033. $\left|x^{2}-5 x\right|<6$.
Solve the inequality $\left|x^{2}-5 x\right|<6$. | ## Solution.
Using the geometric meaning of the modulus, we get that
$$
-6 < x < 6 \quad \text{and} \quad (x+1)(x-6) < 0
$$
This can be written as:
$$
\left\{\begin{array}{l}
(x+1)(x-6) < 0 \\
(x-2)(x-3) > 0
\end{array} \right.
$$
Using the number line, we get \( x \in (-1 ; 2) \cup (3 ; 6) \).
\cup(3;6) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,077 |
9.034. $5 x-20 \leq x^{2} \leq 8 x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.034. $5 x-20 \leq x^{2} \leq 8 x$. | ## Solution.
Let's rewrite this inequality as a system of inequalities
$$
\left\{\begin{array}{l}
x^{2}-5 x+20 \geq 0 \\
x^{2}-8 x \leq 0
\end{array}\right.
$$
Since $x^{2}-5 x+20>0$ for $x \in R$, the given inequality is equivalent to the inequality $x^{2}-8 x \leq 0, x(x-8) \leq 0$. Using the interval method, we f... | x\in[0;8] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,078 |
9.035. $\frac{4 x^{2}-1}{\log _{1.7}\left(\frac{1}{2}\left(1-\log _{7} 3\right)\right)} \leq 0$. | Solution.
$\log _{1.7}\left(\frac{1}{2}\left(1-\log _{7} 3\right)\right)<0$, so the original inequality is equivalent to the following: $4 x^{2}-1 \geq 0$. Solving it, we find $x \geq \frac{1}{2}$ or $x \leq-\frac{1}{2}$.
Answer: $\quad x \in\left(-\infty ;-\frac{1}{2}\right] \cup\left[\frac{1}{2} ; \infty\right)$ | x\in(-\infty;-\frac{1}{2}]\cup[\frac{1}{2};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,079 |
9.039. $\frac{\log _{5}\left(x^{2}+3\right)}{4 x^{2}-16 x}<0$. | ## Solution.
$\log _{5}\left(x^{2}+3\right)>0$ for $x \in R$, therefore $4 x^{2}-16 x<0, x(x-4)<0$.
Answer: $\quad x \in(0 ; 4)$. | x\in(0;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,082 |
9.040. $\frac{x-7}{\sqrt{4 x^{2}-19 x+12}}<0$.
Solve the inequality $\frac{x-7}{\sqrt{4 x^{2}-19 x+12}}<0$. | ## Solution.
The given inequality is equivalent to the system of inequalities
$$
\left\{\begin{array}{l}
x-70
\end{array}\right.
$$
Using the interval method, we obtain \( x \in \left(-\infty ; \frac{3}{4}\right) \cup (4 ; 7) \).
\cup(4;7) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,083 |
9.041. $x^{6}-9 x^{3}+8>0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
9.041. $x^{6}-9 x^{3}+8>0$. | Solution.
Let $x^{3}=y$. Then we get $y^{2}-9 y+8>0$. Solving this inequality, we find $\left[\begin{array}{l}y>8, \\ y<1.\end{array}\right.$
Since $y = x^3$, we have $\left[\begin{array}{l}x^{3}>8, \\ x^{3}<1.\end{array}\right.$
This can be rewritten as $\left[\begin{array}{l}x>2, \\ x<1.\end{array}\right.$
Since ... | x\in(-\infty;1)\cup(2;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,084 |
9.042. $0.3^{2+4+6+\ldots+2x}>0.3^{72}(x \in N)$. | ## Solution.
The given inequality is equivalent to the inequality $2+4+6+\ldots+2x<72$, where the left side of the inequality is the sum of the terms of an arithmetic progression with $a_{1}=2, d=2, a_{n}=2x$. Then we get
$$
\left\{\begin{array}{l}
\frac{2+2x}{2} \cdot n < 72 \\
n = \frac{2x-2}{2} + 1
\end{array} \Le... | 1,2,3,4,5,6,7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,085 |
9.043. $\sqrt{x^{2}-x-12}<x$. | ## Solution.
The given inequality is equivalent to the system of inequalities
$$
\left\{\begin{array} { l }
{ x ^ { 2 } - x - 12 \geq 0 , } \\
{ x > 0 , } \\
{ x ^ { 2 } - x - 12 > 0 } \\
{ x > -12 }
\end{array}\right.} \\
\text { is }
\end{array}\right.\right.
$$
Using the number line, we get $x \in [4 ; \infty)$.... | x\in[4;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,086 |
9.044. $\frac{\sqrt{17-15 x-2 x^{2}}}{x+3}>0$.
9.044. $\frac{\sqrt{17-15 x-2 x^{2}}}{x+3}>0$. | ## Solution.
The given inequality is equivalent to a system of two inequalities
$$
\left\{\begin{array} { l }
{ 17 - 15 x - 2 x ^ { 2 } > 0 , } \\
{ x + 3 > 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 2 x ^ { 2 } + 15 x - 17 < 0 , } \\
{ x > -3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
... | x\in(-3;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,087 |
9.047. $\frac{x^{4}+x^{2}+1}{x^{2}-4 x-5}<0$.
9.047. $\frac{x^{4}+x^{2}+1}{x^{2}-4 x-5}<0$. | ## Solution.
Since $x^{4}+x^{2}+1>0$ for $x \in R$, we have $x^{2}-4 x-5<0$, $(x+1)(x-5)<0, -1<x<5$.
Answer: $\quad x \in(-1 ; 5)$. | x\in(-1;5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,090 |
9.048. $\frac{4-x}{x-5}>\frac{1}{1-x}$. | ## Solution.
Domain of definition: $\quad x \neq 1, x \neq 5$.
From the condition
$$
\begin{aligned}
& \frac{4-x}{x-5}-\frac{1}{1-x}>0, \frac{(4-x)(1-x)-x+5}{(x-5)(1-x)}>0, \frac{x^{2}-6 x+9}{(x-5)(x-1)}<0 \\
& \frac{(x-3)^{2}}{(x-5)(x-1)}<0,(x-3)^{2}(x-5)(x-1)<0
\end{aligned}
$$
Using the interval method, we find ... | x\in(1;3)\cup(3;5) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,091 |
9.049. $\lg 10^{\lg \left(x^{2}+21\right)}>1+\lg x$. | Solution.
Domain of definition: $x>0$.
We get
$\lg \left(x^{2}+21\right)>\lg 10+\lg x \Leftrightarrow \lg \left(x^{2}+21\right)>\lg 10 x \Leftrightarrow\left\{\begin{array}{l}x^{2}+21>10 x, \\ x>0\end{array} \Leftrightarrow\right.$
$\Leftrightarrow\left\{\begin{array}{l}x^{2}-10 x+21>0, \\ x>0\end{array} \Leftright... | x\in(0;3)\cup(7;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,092 |
9.050. $\frac{x^{2}-3 x+2}{x^{2}+3 x+2} \geq 1$.
Solve the inequality:
9.050. $\frac{x^{2}-3 x+2}{x^{2}+3 x+2} \geq 1$. | Solution.
Domain of definition: $x \neq -2, x \neq -1$.
We have
$\frac{x^{2}-3 x+2}{x^{2}+3 x+2}-1 \geq 0 \Leftrightarrow \frac{-6 x}{x^{2}+3 x+2} \geq 0, \frac{x}{(x+2)(x+1)} \leq 0,\left\{\begin{array}{l}x(x+2)(x+1) \leq 0, \\ x \neq -2, \\ x \neq -1 .\end{array}\right.$
Using the interval method, we find $x \in(... | x\in(-\infty;-2)\cup(-1;0] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,093 |
9.053. $x^{2} \cdot 3^{x}-3^{x+1} \leq 0$. | Solution.
From the condition $x^{2} \cdot 3^{x}-3 \cdot 3^{x} \leq 0,3^{x}\left(x^{2}-3\right) \leq 0$. Since $3^{x}>0$ for $x \in R$, the obtained inequality is equivalent to the inequality $x^{2}-3 \leq 0$, $x^{2} \leq 3,-\sqrt{3} \leq x \leq \sqrt{3}$.
Answer: $\quad x \in[-\sqrt{3} ; \sqrt{3}]$ | x\in[-\sqrt{3};\sqrt{3}] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,095 |
9.054. $5^{2 x+1}>5^{x}+4$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.054. $5^{2 x+1}>5^{x}+4$. | Solution.
We have $5 \cdot\left(5^{x}\right)^{2}-\left(5^{x}\right)+4>0$. Solving it as a quadratic equation in terms of $5^{x}$, we get $5^{x} > 1$, from which $x > 0$.
Answer: $\quad x \in (0 ; \infty)$ | x\in(0;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,096 |
9.059. $4^{x}-2^{2(x-1)}+8^{\frac{2}{3}(x-2)}>52$. | ## Solution.
## Given
$4^{x}-4^{x-1}+4^{x-2}>52, 4^{x}-\frac{4^{x}}{4}+\frac{4^{x}}{16}>52, 13 \cdot 4^{x}>16 \cdot 52$,
$4^{x}>4^{3} \Leftrightarrow x>3$.
Answer: $\quad x \in(3 ; \infty)$. | x\in(3;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,098 |
9.060. $2 \log _{8}(x-2)-\log _{8}(x-3)>\frac{2}{3}$. | ## Solution.
## Domain of the function: $\quad x>3$.
## We have
$$
\log _{8}(x-2)^{2}-\log _{8}(x-3)>\frac{2}{3}, \log _{8} \frac{(x-2)^{2}}{x-3}>\frac{2}{3} \Leftrightarrow\left\{\begin{array}{l}
\frac{(x-2)^{2}}{x-3}>8^{\frac{2}{3}} \\
x>3
\end{array} \Leftrightarrow\right.
$$
$$
\Leftrightarrow\left\{\begin{arra... | x\in(3;4)\cup(4;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,099 |
9.061. $25^{x}<6 \cdot 5^{x}-5$. | ## Solution.
By writing the inequality as $\left(5^{x}\right)^{2}-6 \cdot\left(5^{x}\right)+5<0$ and solving it as a quadratic equation in terms of $5^{x}$, we get $1<5^{x}<5$, from which we find $0<x<1$.
Answer: $\quad x \in(0 ; 1)$ | x\in(0;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,100 |
9.062. $\left(\frac{2}{5}\right)^{\log _{0.25}\left(x^{2}-5 x+8\right)} \leq 2.5$. | ## Solution.
Domain of definition: $x \in R$.
## We have
$\left(\frac{2}{5}\right)^{\log _{0.25}\left(x^{2}-5 x+8\right)} \leq\left(\frac{2}{5}\right)^{-1}, \log _{0.25}\left(x^{2}-5 x+8\right) \geq-1 \Leftrightarrow x^{2}-5 x+8 \leq 4 \Leftrightarrow$
$\Leftrightarrow x^{2}-5 x+4 \leq 0,1 \leq x \leq 4$.
Answer: ... | x\in[1;4] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,101 |
9.063. $4^{\frac{1}{x}-1}-2^{\frac{1}{x}-2}-3 \leq 0$. | Solution.
Domain of definition: $x \neq 0$.
From the condition $\frac{\left(2^{\frac{1}{x}}\right)^{2}}{4}-\frac{2^{\frac{1}{x}}}{4}-3 \leq 0 \Leftrightarrow\left(2^{\frac{1}{x}}\right)^{2}-2^{\frac{1}{x}}-12 \leq 0$, solve it as a quadratic equation in terms of $2^{\frac{1}{x}}$. We get $2^{\frac{1}{x}} \geq-3$, fro... | x\in(-\infty;0)\cup[\frac{1}{2};\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,102 |
9.064. $\left(\frac{2}{7}\right)^{3(2 x-7)} \cdot 12,25^{\frac{4 x+1}{2}} \geq 1$. | Solution.
From the condition
$$
\begin{aligned}
& \left(\frac{2}{7}\right)^{3(2 x-7)} \cdot\left(\frac{49}{4}\right)^{\frac{4 x+1}{2}} \geq 1 \Leftrightarrow\left(\frac{2}{7}\right)^{3(2 x-7)} \geq\left(\frac{2}{7}\right)^{4 x+1} \Leftrightarrow \\
& \Leftrightarrow 3 \cdot(2 x-7) \leq 4 x+1, x \leq 11
\end{aligned}
... | x\in(-\infty;11] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,103 |
9.065. $\frac{15}{4+3 x-x^{2}}>1$.
9.065. $\frac{15}{4+3 x-x^{2}}>1$. | ## Solution.
Let's move 1 to the left side of the inequality and bring it to a common denominator. We have $\frac{x^{2}-3 x+11}{x^{2}-3 x-4} - 1 = \frac{x^{2}-3 x+11 - (x^{2}-3 x-4)}{x^{2}-3 x-4} = \frac{15}{x^{2}-3 x-4}$. Since $x^{2}-3 x-4 \neq 0$ for $x \in R$, then $x^{2}-3 x-4 < 0$. Solving this inequality, we fi... | x\in(-1;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,104 |
9.070. $\log _{1.5} \frac{2 x-8}{x-2}<0$. | Solution.
The given inequality is equivalent to a system of two inequalities
$\left\{\begin{array}{l}\frac{2 x-8}{x-2}0\end{array} \Leftrightarrow\left\{\begin{array}{l}\frac{x-6}{x-2}0\end{array} \Leftrightarrow\left\{\begin{array}{l}(x-6)(x-2)0 .\end{array}\right.\right.\right.$
Using the number line, we get $x \i... | x\in(4;6) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,107 |
9.071. $\log _{0.3}\left(x^{2}-5 x+7\right)>0$. | ## Solution.
The given inequality is equivalent to the system of two inequalities
$\left\{\begin{array}{l}x^{2}-5 x+7>0, \\ x^{2}-5 x+7<1\end{array} \Leftrightarrow\left\{\begin{array}{l}x \in R, \\ 2<x<3,\end{array} \quad 2<x<3\right.\right.$.
Answer: $\quad x \in(2 ; 3)$. | x\in(2;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,108 |
9.072. $x^{8}-6 x^{7}+9 x^{6}-x^{2}+6 x-9<0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.072. $x^{8}-6 x^{7}+9 x^{6}-x^{2}+6 x-9<0$. | Solution.
Rewrite the inequality as
$$
\begin{aligned}
& x^{6}\left(x^{2}-6 x+9\right)-\left(x^{2}-6 x+9\right)0 \quad \text{and} \quad x^{2}-x+1>0 \quad \text{for} \quad x \in \mathbb{R}, \text{so it is equivalent to the inequality} \quad (x-3)^{2}(x-1)(x+1)<0. \text{Using the interval method, we get} \quad x \in(-1... | x\in(-1;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,109 |
9.073. $a^{4}+a^{3}-a-1<0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.073. $a^{4}+a^{3}-a-1<0$. | Solution.
Grouping, we get
$$
a^{3}(a+1)-(a+1)0$ for $a \in R$, so it is equivalent to the inequality $(a+1)(a-1)<0$. Using the number line, we get $a \in(-1 ; 1)$.
Answer: $\quad a \in(... | \in(-1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,110 |
9.074. $m^{3}+m^{2}-m-1>0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.074. $m^{3}+m^{2}-m-1>0$. | Solution.
From the condition
$$
\begin{aligned}
& m^{2}(m+1)-(m+1)>0,(m+1)\left(m^{2}-1\right)>0,(m+1)(m+1)(m-1)>0 \\
& (m+1)^{2}(m-1)>0 \Leftrightarrow\left\{\begin{array} { l }
{ m - 1 > 0 , } \\
{ m + 1 \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
m>1, \\
m \neq-1,
\end{array} \quad m>1\right.\rig... | \in(1;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,111 |
9.075. $\log _{2}\left(1+\log _{\frac{1}{9}} x-\log _{9} x\right)<1$. | Solution.
The given inequality is equivalent to the double inequality
$0<1+\log _{\frac{1}{9}} x-\log _{9} x<2, -1<\log _{\frac{1}{9}} x-\log _{9} x<1$.
We will change the base to $\frac{1}{9}$. We have
$-1<\log _{\frac{1}{9}} x+\log _{\frac{1}{9}} x<1, -1<2 \log _{\frac{1}{9}} x<1, -\frac{1}{2}<\log _{\frac{1}{9}}... | x\in(\frac{1}{3};3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,112 |
9.076. $\sqrt{x^{\log _{2} \sqrt{x}}}>2$.
9.076. $\sqrt{x^{\log _{2} \sqrt{x}}}>2$. | ## Solution.
Domain of definition: $x>0$.
From the condition $x^{\frac{\log _{2} \sqrt{x}}{2}}>2, x^{\frac{\log _{2} x}{4}}>2$. Taking the logarithm of both sides of the inequality with base 2, we get
$\log _{2} x^{\frac{\log _{2} x}{4}}>\log _{2} 2, \frac{\log _{2} x}{4} \cdot \log _{2} x>1, \log _{2}^{2} x>4 \Leftr... | x\in(0;\frac{1}{4})\cup(4;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,113 |
9.077. $2^{x+2}-2^{x+3}-2^{x+4}>5^{x+1}-5^{x+2}$. | Solution.
We have
$$
\begin{aligned}
& 4 \cdot 2^{x}-8 \cdot 2^{x}-16 \cdot 2^{x}>5 \cdot 5^{x}-25 \cdot 5^{x},-20 \cdot 2^{x}>-20 \cdot 5^{x} \Leftrightarrow \\
& \Leftrightarrow\left(\frac{2}{5}\right)^{x}<1
\end{aligned}
$$
Answer: $\quad x \in(0 ; \infty)$. | x\in(0;\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,114 |
9.079. $\frac{x^{3}-x^{2}+x-1}{x+8} \leq 0$.
9.079. $\frac{x^{3}-x^{2}+x-1}{x+8} \leq 0$. | ## Solution.
Domain of definition: $x \neq-8$.
Rewrite the inequality as $\frac{x^{2}(x-1)+(x-1)}{x+8} \leq 0, \frac{(x-1)\left(x^{2}+1\right)}{x+8} \leq 0$. Since $x^{2}+1>0$ for $x \in R$, the obtained inequality is equivalent to the system of inequalities $\left\{\begin{array}{l}(x-1)(x+8) \leq 0, \\ x \neq-8 .\en... | x\in(-8;1] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,116 |
9.080. $\frac{x^{4}-2 x^{2}-8}{x^{2}+2 x+1}<0$.
9.080. $\frac{x^{4}-2 x^{2}-8}{x^{2}+2 x+1}<0$. | Solution.
Domain of definition: $\quad x \neq-1$.
By solving the biquadratic equation $x^{4}-2 x^{2}-8=0$, we can represent the inequality as $\frac{\left(x^{2}+2\right)\left(x^{2}-4\right)}{(x+1)^{2}}0$ for $x \in R$ and
$(x+1)^{2}>0$ for $x \neq-1$, then this inequality is equivalent to the system of inequalities $... | x\in(-2;-1)\cup(-1;2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,117 |
9.081. $\log _{1.2}(x-2)+\log _{1.2}(x+2)<\log _{1.2} 5$. | ## Solution.
Domain of definition: $x>2$.
We have $\log _{1.2}(x-2)(x+2) 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ x ^ { 2 } 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
-32
\end{array} \Leftrightarrow 2<x<3\right.\right.\right.
$
Answer: $\quad x \in(2 ; 3)$ | x\in(2;3) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,118 |
9.082. $\frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}>1$. | ## Solution.
Domain of definition: $x \neq-3, x \neq-2, x \neq-1$.
We have
$$
\begin{aligned}
& \frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}-1>0 \Leftrightarrow \frac{-12 x^{2}-6}{(x+1)(x+2)(x+3)}>0 \Leftrightarrow \\
& \Leftrightarrow \frac{2 x^{2}+1}{(x+1)(x+2)(x+3)}<0 \text{. Since } 2 x^{2}+1>0 \text{ for } x \in \ma... | x\in(-\infty;-3)\cup(-2;-1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,119 |
9.083. $\frac{1}{3^{x}+5}<\frac{1}{3^{x+1}-1}$. | ## Solution.
Domain of definition: $x \neq-1$.
From the condition
$$
\frac{1}{3^{x}+5}-\frac{1}{3 \cdot 3^{x}-1}<0
$$
Since $3^{x}+5>0$ and $3 \cdot 3^{x}-1>0$ for $x \in \mathbb{R}$, this inequality is equivalent to the inequality $\frac{2 \cdot 3^{x}-6}{3 \cdot 3^{x}-1}<0,\left(3^{x}-3\right)\left(3^{x}-\frac{1}{... | x\in(-1;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,120 |
9.084. $\log _{x}\left(\log _{9}\left(3^{x}-9\right)\right)<1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \in(0 ; 1) \cup(1 ;+\infty), \\ 3^{x}-9>1,\end{array}\left\{\begin{array}{l}x \in(0 ; 1) \cup(1 ;+\infty) \\ 3^{x}>10,\end{array} x \in\left(\log _{3} 10 ;+\infty\right)\right.\right.$.
We obtain the following system of inequalities
$$
\begin{aligned}
& \l... | x\in(\frac{1}{\lg3};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,121 |
9.085. $0.2^{\frac{x^{2}+2}{x^{2}-1}}>25$. | Solution.
Domain of definition: $\quad x \neq \pm 1$.
Rewrite the inequality as
$5^{-\frac{x^{2}+2}{x^{2}-1}}>5^{2} \Leftrightarrow-\frac{x^{2}+2}{x^{2}-1}>2, \frac{x^{2}+2}{x^{2}-1}+2<0, \frac{x^{2}+2+2 x^{2}-2}{x^{2}-1}<0$,
$\frac{3 x^{2}}{x^{2}-1}<0,\left\{\begin{array}{l}x \neq 0, \\ x^{2}-1<0\end{array} \Leftr... | x\in(-1;0)\cup(0;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,122 |
9.086. $5^{2 \sqrt{x}}+5<5^{\sqrt{x}+1}+5^{\sqrt{x}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
9.086. $5^{2 \sqrt{x}}+5<5^{\sqrt{x}+1}+5^{\sqrt{x}}$. | Solution.
Let's represent the given inequality as $\left(5^{\sqrt{x}}\right)^{2}+5<5 \cdot 5^{\sqrt{x}}+5^{\sqrt{x}}$, $\left(5^{\sqrt{x}}\right)^{2}-6 \cdot 5^{\sqrt{x}}+5<0$. Solving this inequality as a quadratic equation in terms of $5^{\sqrt{x}}$, we get $1<5^{\sqrt{x}}<5, \quad 0<\sqrt{x}<1, \quad 0<x<1$.
Answe... | x\in(0;1) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,123 |
9.087. $\left|3-\log _{2} x\right|<2$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
9.087. $\left|3-\log _{2} x\right|<2$. | ## Solution.
Domain: $\quad x>0$.
The given inequality is equivalent to the inequality
$-2<3-\log _{2} x<2,-5<-\log _{2} x<-1 \Leftrightarrow 1<\log _{2} x<5,2<x<32$.
Answer: $\quad x \in(2 ; 32)$. | x\in(2;32) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 51,124 |
9.090. $3^{\sqrt{x}}+3^{\sqrt{x}-1}-3^{\sqrt{x}-2}<11$. | Solution.
Domain of definition: $x \geq 0$.
From the condition $3^{\sqrt{x}}+\frac{3^{\sqrt{x}}}{3}-\frac{3^{\sqrt{x}}}{9}<11.11 \cdot 3^{\sqrt{x}}<9 \cdot 11,3^{\sqrt{x}}<3^{2}$. This is equivalent to the inequality $\sqrt{x}<2,0 \leq x<4$.
Answer: $\quad x \in [0 ; 4)$. | x\in[0;4) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,126 |
9.092. $\log _{0.5}^{2} x+\log _{0.5} x-2 \leq 0$. | Solution.
Domain of definition: $\quad x>0$.
Solving this inequality as a quadratic inequality with respect to $\log _{0,5} x$, we get $-2 \leq \log _{0,5} \cdot x \leq 1,0,5 \leq x \leq 4$.
Answer: $\quad x \in[0,5 ; 4]$ | x\in[0.5;4] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,127 |
9.095. $\log _{4}(x+7)>\log _{2}(x+1)$. | Solution.
Domain of definition: $x>-1$.
Let's switch to base 2. We have
$$
\begin{aligned}
& \frac{1}{2} \log _{2}(x+7)>\log _{2}(x+1) \Leftrightarrow \log _{2}(x+7)>2 \log _{2}(x+1) \Leftrightarrow \\
& \Leftrightarrow \log _{2}(x+7)>\log _{2}(x+1)^{2}
\end{aligned}
$$
The last inequality is equivalent to the syst... | x\in(-1;2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 51,128 |
10.002. Find the diagonal and the lateral side of an isosceles trapezoid with bases 20 and 12 cm, given that the center of the circumscribed circle lies on the larger base of the trapezoid. | ## Solution.
Since $A D$ is the diameter of the circle (Fig. 10.2), then $O D = O C = 10$ cm. Draw $C L \perp A D$; then $O L = 6$ cm and from $\triangle C L O$ we find
$$
C L = \sqrt{O C^{2} - O L^{2}} = 8 \, \text{cm}
$$
Then from $\triangle A L C$ and $\triangle C L D$ we have
$$
A C = \sqrt{C L^{2} + A L^{2}} =... | 4\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,129 |
10.004. The height of the rhombus, drawn from the vertex of the obtuse angle, divides its side into segments of length $m$ and $n$. Determine the diagonals of the rhombus. | Solution.
The length of the side of the rhombus is $m+n$. From $\triangle A B K$ (Fig. 10.4) we find $B K^{2}=(m+n)^{2}-m^{2}$. In $\triangle B K D$ we have
$$
B D^{2}=B K^{2}+n^{2}=(m+n)^{2}-m^{2}+n^{2}=2 n(m+n)
$$
i.e., $B D=\sqrt{2 n(m+n)}$.
},AC=\sqrt{4^{2}+6n+2n^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,130 |
10.005. A square is inscribed in a right triangle with legs $a$ and $b$, sharing a right angle with the triangle. Find the perimeter of the square. | ## Solution.
Let the side of the square be denoted by $x$ (Fig. 10.5). Since $\triangle A C B \sim \triangle F E B$, we have $\frac{a}{x}=\frac{b}{b-x}$. Therefore,
$$
x=\frac{a b}{a+b}, P=4 x=\frac{4 a b}{a+b}
$$
Answer: $\frac{4 a b}{a+b}$. | \frac{4}{+b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,131 |
10.006. Two circles with radii $R=3$ cm and $r=1$ cm touch each other externally. Find the distance from the point of tangency of the circles to their common tangents. | Solution.
Let $F M=x$ (Fig. 10.6). From $\triangle O_{1} T O \sim \Delta F M O$ we have $\frac{R-r}{x}=\frac{R+r}{r},$ hence $x=\frac{r(R-r)}{R+r},$ and $E F=x+r=\frac{r(R-r)}{R+r}+r=\frac{2 R r}{R+r}=$ $=\frac{6}{4}=\frac{3}{2}$ cm. If, however, the common tangent of the circles is considered to be the line $E F$, th... | \frac{3}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,132 |
10.007. An isosceles trapezoid with a side length of 17 cm is circumscribed around a circle with a diameter of 15 cm. Find the bases of the trapezoid. | Solution.
According to the condition, $K Q=15 \mathrm{~cm}, A B=C D=17 \mathrm{~cm}, B C+A D=A B+C D=34$. In $\triangle C E D \angle C E D=90^{\circ}, C E=K Q$ (Fig. 10.7) We have $D E=\sqrt{289-225}=8 \mathrm{~cm}$, $A D=2 E D+B C=16+B C, B C+16+B C=34, B C=9 \mathrm{~cm}, A D=25 \mathrm{~cm}$.
Answer: 9 cm $; 25$ c... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,133 |
10.008. In an isosceles triangle with a lateral side equal to 4 cm, a median of the lateral side is drawn. Find the base of the triangle if the median is 3 cm. | Solution.
Using the cosine theorem, we get $\left\{\begin{array}{l}x^{2}=16+16-2 \cdot 16 \cos \alpha, \\ m^{2}=16+4-2 \cdot 8 \cos \alpha\end{array} \Rightarrow\right.$
$\Rightarrow x^{2}-2 m^{2}=-8, x^{2}=10, x=\sqrt{10}$ (Fig. 10.8).
Answer: $\sqrt{10}$ cm. | \sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,134 |
10.009. In an isosceles triangle, the base is 16 cm, and the lateral side is $10 \mathrm{~cm}$. Find the radii of the inscribed and circumscribed circles and the distance between their centers. | Solution.
$$
S_{\triangle A B C}=\frac{1}{2} A C \cdot B D, \text { where } B D=\sqrt{A B^{2}-A D^{2}}=\sqrt{10^{2}-8^{2}}=6 \text { (Fig.10.9), }
$$
i.e., $S_{\triangle A B C}=6 \cdot 8=48\left(\mathrm{~cm}^{2}\right)$. Let $R$ and $r$ be the radii of the circumscribed and inscribed circles of the triangle. Then
, so $\angle A B O_{1}=30^{\circ}$. From $\triangle B A O_{1}$ we have $B O_{1}=\frac{R}{\sin 30^{\circ}}=2 R$, from $\triangle B E O_{2} \angle B E O_{2}=90... | proof | Geometry | proof | Yes | Yes | olympiads | false | 51,139 |
10.016. In the sector $A O B$ with radius $R$ and angle $90^{\circ}$, a circle is inscribed, touching segments $O A, O B$ and arc $A B$ (Fig. 10.16). Find the radius of the circle. | ## Solution.
$\triangle O M F$ is isosceles $\left(\angle M O B=\angle M F O=45^{\circ}\right)$. Therefore, $M O=M F$, $O F=R \sqrt{2}$. From $\triangle O_{1} Q O \sim \Delta O M F$ we have $\frac{r}{R}=\frac{R-r}{\sqrt{2} R}, r=R(\sqrt{2}-1)$.
Answer: $\quad R(\sqrt{2}-1)$
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,140 |
10.017. Given a point $P$, which is 7 cm away from the center of a circle with a radius of $11 \mathrm{~cm}$. A chord of length 18 cm is drawn through this point. What are the lengths of the segments into which the chord is divided by point $P$? | ## Solution.
Draw the diameter $C D$ through point $P$ (Fig. 10.17), which will divide it into segments $P D$ and $C P$ of lengths $11-7=4$ and $11+7=18$ (cm). Let $A P=x$; then $P B=18-x$. Since $A P \cdot P B=C P \cdot P D=4 \cdot 18$, we have $x(18-x)=72$ or $x^{2}-18 x+72=0$, from which $x_{1}=12, x_{2}=6$, i.e., ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,141 |
10.018. Find the lengths of sides $A B$ and $A C$ of triangle $A B C$, if $B C=8 \mathrm{~cm}$, and the lengths of the heights drawn to $A C$ and $B C$ are 6.4 and 4 cm, respectively. | Solution.
According to the problem, $B C=8 \text{ cm}, A E=4$ cm, $B K=6.4$ cm (Fig. 10.18). From $\triangle B K C \sim \triangle A E C$, it follows that $\frac{6.4}{4}=\frac{8}{A C} ; A C=5$ cm. From $\triangle A E C$, we have $C E=\sqrt{25-16}=3 \text{ cm}, B E=8-3=5 \text{ cm}$; then from $\triangle A E B$, we find... | \sqrt{41} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,142 |
10.020. A rhombus with diagonals of 12 and 6 cm is inscribed in the intersection of two equal circles. Find the radius of the circle. | Solution.
According to the condition, $ABCD$ is a rhombus, $BD=12$ cm, $AC=6$ cm (Fig. 10.20). Let $O_1O=x, R$ be the radius of the circles. Then $x=R-\frac{AC}{2}=R-3$. In $\triangle O_1OB \quad \angle O_1OB=90^{\circ}, R^2=36+(R-3)^2, R=\frac{45}{6}=\frac{15}{2} \text{ cm}$.
Answer: $7.5 \text{ cm}$. | 7.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,143 |
10.021. The median to the hypotenuse of a right triangle is equal to $m$ and divides the right angle in the ratio 1:2. Find the sides of the triangle. | Solution.
By the condition $C O=m, A O=B O, \frac{\angle B C O}{\angle A C O}=\frac{2}{1}, A B=2 m(O-$ is the center of the circumscribed circle) (Fig. 10.21). Let $\angle A C O=x, \angle B C O=2 x$, $x+2 x=90^{\circ}, x=30^{\circ}, \quad \angle A C O=\angle C A O . B \quad \triangle A B C \quad \angle A C B=90^{\circ... | ;\sqrt{3},2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,144 |
10.022. Determine the acute angles of a right triangle if the median drawn to its hypotenuse divides the right angle in the ratio 1:2. | Solution.
$A O=O B, \frac{\angle B C O}{\angle A C O}=\frac{2}{1}$ (Fig. 10.22). In $\triangle A B C, \angle A C B=90^{\circ}$; thus, $O$ is the center of the circumscribed circle, meaning $A O=O B=C O$, $\triangle A O C$ is isosceles, $\angle A=\angle O C A, \angle B=\angle B C O$. Let
, in $\triangle O E A ~ \angle O E A=90^{\circ}$, $E A^{2}=O A^{2}-O E^{2}, R^{2}-(x-R)^{2}=\frac{x^{2}}{4}, \frac{x^{2}}{4}+x^{2}=2 x R ; x=\frac{8 R}{5} ; A C=x \sqrt{2=}$ $=\frac{8 R}{5} \cdot \sqrt{2}=1.6 R \sqrt{2}$.
Answer: $1.6 R \sqrt{2}$. | 1.6R\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,146 |
10.024. The lengths of the parallel sides of the trapezoid are 25 and 4 cm, and the lengths of the non-parallel sides are 20 and 13 cm. Find the height of the trapezoid. | ## Solution.
Given $B C=4 \text{ cm}, A D=25 \text{ cm}, A B=20 \text{ cm}, C D=13 \text{ cm}$ (Fig. 10.24). Draw $B E \perp A D$ and $C F \perp A D$. Let $B E=C F=h, A E=x$, $F D=y$. Then from $\triangle A B E$ and $\triangle C F D$ we find $h^{2}=20^{2}-x^{2}=13^{2}-y^{2}$. Considering that $y=25-4-x=21-x$, we have ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,147 |
10.027. Each of the three equal circles of radius $r$ touches the other two. Find the area of the triangle formed by the common external tangents to these circles. | ## Solution.
Triangle $A B C$ is equilateral (Fig. 10.27); therefore, its area is $\frac{a^{2} \sqrt{3}}{4}$, where $a$ is the side of the triangle. Draw $O K \perp A C$. Since $\angle O A K=30^{\circ}$, then $A O=2 r, A K=r \sqrt{3}$, hence $a=2 r \sqrt{3}+2 r=2 r(\sqrt{3}+1)$. Consequently, $S=\frac{4 r^{2}(3+2 \sqr... | 2r^{2}(2\sqrt{3}+3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,149 |
10.028. The bases of an isosceles trapezoid are $a$ and $b$, its lateral side is $c$, and its diagonal is $d$. Prove that $d^{2}=a b+c^{2}$. | Solution.
Given $B C=a, A C=d, C D=c, A D=b$ (Fig. 10.28). In $\triangle C E D$
$a$
. | Solution.
1st case. By the condition $A B=a=A N=N M=M B=B D=A D, O_{1} E=\frac{a}{2}$, $\angle B O_{2} E=2 \angle E D B=60^{\circ}, O_{2} E=E B \cdot \operatorname{ctg} 60^{\circ}=\frac{a}{2 \sqrt{3}}$,
$$
O_{1} O_{2}=\frac{a}{2 \sqrt{3}}+\frac{a}{2}=\frac{a}{6}(3+\sqrt{3})(\text { fig. } 10.29, a) .
$$
2nd case. $O... | \frac{(3+\sqrt{3})}{6},\frac{(3-\sqrt{3})}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,151 |
10.030. On the sides of a square, equilateral triangles are constructed outside it, and their vertices are sequentially connected. Determine the ratio of the perimeter of the resulting quadrilateral to the perimeter of the given square. | Solution.
Let the side of the square be $x$, then the perimeter of the square
$P_{0}=4 x, \triangle F B M$ is isosceles, $\angle F B M=360^{\circ}-90^{\circ}-2 \cdot 60^{\circ}=150^{\circ}$... | \frac{\sqrt{6}+\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,152 |
10.031. A circle with a radius of 2 is inscribed in a rhombus, which is divided by its diagonal into two equilateral triangles. Find the side of the rhombus. | ## Solution.
Let $a$ be the side of the rhombus, and $d_{1}$ and $d_{2}$ be its diagonals (Fig. 10.31). Since the height of the rhombus is equal to the diameter of the circle, its area is $a$. On the other hand, $S=\frac{1}{2} d_{1} d_{2}$, from which, considering that $d_{1}=a$, we find $d_{2}=8$. But $a^{2}=\left(\f... | \frac{8\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,153 |
10.033. A tangent, parallel to the base, is drawn to the circle inscribed in an isosceles triangle with a base of 12 cm and a height of 8 cm. Find the length of the segment of this tangent, enclosed between the sides of the triangle. | ## Solution.
Let's find the length of the side $B C$ (Fig. 10.33): $B C=\sqrt{B M^{2}+M C^{2}}=$ $=\sqrt{6^{2}+8^{2}}=10$ (cm). Considering that $A O$ is the bisector of $\triangle A B M$, we have $M O / O B=A M / A B$ or $r /(8-r)=6 / 10$, from which $r=3$ (cm). Since $D E \| A C$, then $\triangle D B E \sim \triangl... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,155 |
10.035. From point $A$, two lines are drawn tangent to a circle of radius $R$ at points $B$ and $C$ such that triangle $A B C$ is equilateral. Find its area (Fig. 10.35). | ## Solution.
Given $O B=O C=R, A B=A C=B C$. Since $\triangle A B C$ is equilateral, then $S=B C^{2} \frac{\sqrt{3}}{4} ; \angle A B O=90^{\circ}, \angle A B C=60^{\circ}, \angle C B O=30^{\circ}$, $B C=2 \cdot R \cdot \cos 30=\sqrt{3} \cdot R, S=3 R^{2} \frac{\sqrt{3}}{4}$.
Answer: $\quad 3 R^{2} \frac{\sqrt{3}}{4}$... | 3R^{2}\frac{\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,156 |
10.036. A rhombus with a side length of 6 cm is inscribed in a right-angled triangle with a $60^{\circ}$ angle, such that the $60^{\circ}$ angle is common to both and all vertices of the rhombus lie on the sides of the triangle. Find the sides of the triangle. | Solution.
In $\triangle F B D$ (Fig. 10.36) the leg $F D=6$ cm and lies opposite the angle of $30^{\circ}$, from which $F B=12$ cm and, therefore, $A B=18$ cm. Next, in $\triangle A B C$ we have $A C=A B / 2=9$ cm. Finally, $B C=\sqrt{A B^{2}-A C^{2}}=\sqrt{243}=9 \sqrt{3}$ cm.
Answer: $9, 9 \sqrt{3}$, and 18 cm.
. Find the area of the triangle and the length of the inscribed semicircle. | Solution.
Given $A O=20 \mathrm{~cm}, O B=15 \mathrm{~cm}$. Let $r$ be the radius of the semicircle; $\triangle A F O \sim \triangle O E B$, we have $\frac{20}{15}=\frac{r}{\sqrt{225-r^{2}}} ; 4 \sqrt{225-r^{2}}=3 r ;$ $16\left(225-r^{2}\right)=9 r^{2}, r^{2}=9 \cdot 16, r=12 \mathrm{~cm} ; l=\pi r=12 \pi$. In $\trian... | 294\mathrm{~}^{2},12\pi\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,161 |
10.041. The measure of one of the angles of a parallelogram is $60^{\circ}$, and the shorter diagonal is $2 \sqrt{31}$ cm. The length of the perpendicular dropped from the point of intersection of the diagonals to the longer side is $\sqrt{75} / 2 \mathrm{~cm}$. Find the lengths of the sides and the longer diagonal of ... | ## Solution.
Draw $B N \perp A D$ (Fig. 10.41); since $B N=2 O M$, then $B N=\sqrt{75}$ cm. Considering that in $\triangle A N B \quad \angle A B N=30^{\circ}$, we have $A B=2 A N$ and, therefore, $4 A N^{2}=75+A N^{2}$, from which $A N=5$ cm and $A B=10$ cm. Next, from $\triangle B N D$ we get $N D^{2}=B D^{2}-B N^{2... | 12;10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,162 |
10.042. One of the angles of the trapezoid is $30^{\circ}$, and the lines containing the lateral sides of the trapezoid intersect at a right angle (Fig. 10.42). Find the length of the shorter lateral side of the trapezoid if its midline is 10 cm, and one of the bases is $8 \mathrm{~cm}$. | ## Solution.
Given $\angle B C A=30^{\circ}, \angle A B C=90^{\circ}, K M=10$ cm, $D E=8$ cm, $\frac{8+A C}{2}=10, A C=12$ cm (since $K M$ is the midline). Since $\triangle D B E \sim \triangle A B C$, then $\frac{A B}{D B}=\frac{A C}{D E} ; \frac{x+D B}{D B}=\frac{12}{8}$ (where $A D=x$ ), $x=D B \cdot \frac{1}{2}$. ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,163 |
10.043. A regular triangle is inscribed in a circle with a diameter equal to $\sqrt{12}$. On its height as a side, another regular triangle is constructed, in which a new circle is inscribed. Find the radius of this circle.
 is $R \sqrt{3}$, i.e., $0.5 \sqrt{12} \cdot \sqrt{3}=3$. We will find the side $CD$ of the new triangle: $a=\sqrt{3^{2}-1.5^{2}}=1.5 \sqrt{3}$. Since the radius $r$ of the circle inscribed in it is $a \sqrt{3} / 6$, we f... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,164 |
10.044. In a circle, two chords $A B=a$ and $A C=b$ are drawn. The length of arc $A C$ is twice the length of arc $A B$. Find the radius of the circle. | Solution.
Let $\angle B O A=\alpha$, then $\angle A C B=\frac{\alpha}{2}, a^{2}=2 R^{2}-2 R^{2} \cdot \cos \alpha$ ( $R$-the radius to be found) (Fig. 10.44). By the Law of Sines $\frac{a}{\sin \frac{\alpha}{2}}=\frac{b}{\sin \alpha}=2 R$, $\sin \alpha=\frac{b}{2 R}, a^{2}=2 R^{2}-2 R^{2} \cdot \frac{1}{2 R} \sqrt{4 R... | \frac{^{2}}{\sqrt{4^{2}-b^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,165 |
10.045. The common chord of two intersecting circles is seen from the centers at angles of $90^{\circ}$ and $60^{\circ}$. Find the radii of the circles if the distance between their centers is $\sqrt{3}+1$. | Solution.
Let $r$ be the radius of one circle, and $R$ be the radius of the other circle. According to the problem, $\angle A O_{1} B=90^{\circ}, \angle A O_{2} B=60^{\circ}, O_{1} O_{2}=\sqrt{3}+1 ; A B=r \sqrt{2} ;$ $O_{1} E=A B / 2 ; O_{1} E=r \frac{\sqrt{2}}{2}, A B=2 R \cdot \sin 30^{\circ}=R\left(\angle A O_{2} ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,166 |
10.046. A circle touches the larger leg of a right triangle, passes through the vertex of the opposite acute angle, and has its center on the hypotenuse of the triangle. What is the radius of the circle if the lengths of the legs are 5 and $12?$ | Solution.
Given $A C=5, C B=12$ (Fig. 10.46). In $\triangle A B C$, $\angle C=90^{\circ}$, $A B=\sqrt{25+144}=13$. We have $\triangle A C B \sim \triangle O D B \Rightarrow \frac{A B}{O B}=\frac{A C}{O D}$. Let $O D=R$ be the radius of the given circle, then $\frac{13}{O B}=\frac{5}{R}$ and $O B=13-R$, so $\frac{13}{1... | \frac{65}{18} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,167 |
10.047. The perimeter of a right-angled triangle $ABC$ ( $\angle C=90^{\circ}$ ) is 72 cm, and the difference between the lengths of the median $CK$ and the altitude $CM$ is 7 cm (Fig. 10.47). Find the length of the hypotenuse. | Solution.
According to the condition $C K - C M = 7 \text{ cm}, P_{A B C} = 72 \text{ cm}, p = \frac{P_{A B C}}{2} = 36$. If the radius of the inscribed circle, then $r = p - A B, S = p \cdot r = p \cdot (p - A B)$. Let $A B = x, K$ be the center of the circumscribed circle, $C K = A K = K B = \frac{A B}{2} = \frac{x}... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,168 |
10.048. Two circles are inscribed in an acute angle of $60^{\circ}$, touching each other externally. The radius of the smaller circle is $r$. Find the radius of the larger circle. | ## Solution.
By the condition $\angle B A C=60^{\circ}, O_{1} C=r$ (Fig. 10.48). In $\triangle A C O_{1} \angle A C O_{1}=90^{\circ}$, $\angle O_{1} A C=30^{\circ} ; A O_{1}=\frac{O_{1} C}{\sin 30^{\circ}}=2 O_{1} C=2 r$. In $\triangle O_{2} E A \quad \angle O_{2} E A=90^{\circ}$; $\Delta O_{2} E A \sim \Delta O_{1} C... | 3r | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,169 |
10.049. A point on the hypotenuse, equidistant from both legs, divides the hypotenuse into segments of length 30 and $40 \mathrm{~cm}$. Find the legs of the triangle. | Solution.
Given $\angle C=90^{\circ}, A D=30 \text{ cm}, B D=40$ cm (Fig. 10.49). Let $A C=x, B C=y$. Since the point equidistant from the sides of an angle lies on its bisector, $\frac{x}{y}=\frac{30}{40}$, i.e., $y=\frac{4 x}{3}$. But $x^{2}+y^{2}=A B^{2}$ or $x^{2}+\frac{16 x^{2}}{9}=70^{2}$, from which $x^{2}=1764... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,170 |
10.050. Find the radius of the circle circumscribed around a right-angled triangle if the radius of the circle inscribed in this triangle is $3 \mathrm{~cm}$, and one of the legs is $10 \mathrm{~cm}$. | Solution.
Draw radii $O D, O M$ and $O N$ to the points of tangency (Fig. 10.50). We have $O D=O M=C M=C D$ and, therefore, $B D=10-3=7$ (cm). But $B D=B N$ and $A N=A M$ (tangents drawn from the same point). Let $A N=x$; then $(x+7)^{2}=(x+3)^{2}+10^{2}$, from which $8 x=60$, i.e., $x=7.5$ (cm). Hence, $A B=14.5$ (cm... | 7.25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,171 |
10.051. Three circles of different radii touch each other pairwise. The segments connecting their centers form a right triangle. Find the radius of the smallest circle, if the radii of the largest and medium circles are 6 and 4 cm. | Solution.
Fig. 10.51
Fig. 10.52
Let $r$ be the radius of the s... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,172 |
10.052. A circle touches one of the legs of an isosceles right triangle and passes through the vertex of the opposite acute angle. Find the radius of the circle if its center lies on the hypotenuse of the triangle, and the leg of the triangle is equal to $a$. | ## Solution.
By the condition $C B=a, A C=C B$ (Fig. 10.52). Therefore, $\angle C A B=\angle C B A=45^{\circ}$ (since $\angle C=90^{\circ}$) $\Rightarrow A B=a \sqrt{2} \cdot \triangle A C B \sim \triangle O E B \Rightarrow \frac{a \sqrt{2}}{a \sqrt{2}-R}=\frac{a}{R}$, $R=\frac{a \sqrt{2}}{\sqrt{2}+1}=a(2-\sqrt{2})$
... | (2-\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,173 |
10.053. In parallelogram $ABCD$, the height drawn from vertex $B$ of the obtuse angle to side $DA$ divides it in the ratio 5:3, counting from vertex $D$ (Fig. 10.53). Find the ratio $AC:BD$, if $AD:AB=2$. | Solution.
Fig. 10.54
Fig. 10.56
Given $\frac{A D}{A B}=2, \f... | 2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,174 |
10.054. Based on an isosceles triangle with a base equal to 8 cm, a circle is constructed on the base as a chord, touching the lateral sides of the triangle (Fig. 10.54). Find the radius of the circle if the length of the height drawn to the base of the triangle is 3 cm. | ## Solution.
Given $A B=B Q, A Q=8 \text{ cm}, B E=3 \text{ cm}. A B=\sqrt{16+9}=5 \text{ cm}$,
$$
\begin{aligned}
& A E^{2}=B E \cdot O E \quad \Rightarrow \quad O E \cdot 3=16, \quad O E=\frac{16}{3} \cdot \quad A B^{2}=B D \cdot B C \\
& 25=\left(R+3+\frac{16}{3}\right) \cdot\left(\frac{25}{3}-R\right), R^{2}=\fra... | \frac{20}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,175 |
10.055. A circle is inscribed in an isosceles triangle with a vertex angle of $120^{\circ}$ and a lateral side of $a$ (Fig. 10.55). Find the radius of this circle. | Solution.
Given $A B=B C=a, \angle A B C=120^{\circ}, \angle B A C=\angle A C B=30^{\circ}$, $B E=a \sin 30^{\circ}=\frac{a}{2}, S=r \cdot \frac{A B+B C+A C}{2}, A C=2 a \cdot \cos 30^{\circ}=\frac{2 a \sqrt{3}}{2}=a \sqrt{3} ;$ $S=\frac{1}{2} a^{2} \cdot \sin 120^{\circ}=\frac{a^{2} \sqrt{3}}{4} \Rightarrow \frac{a^{... | \frac{\sqrt{3}(2-\sqrt{3})}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,176 |
10.056. Prove that the sum of the distances from any point taken inside a regular polygon to all the lines containing its sides is a constant value. | Solution.
Connect any internal point $P$ with all vertices of the polygon and drop perpendiculars to all sides (or their extensions). Let the lengths of these perpendiculars be $d_{1}, d_{2}, \ldots, d_{n}$. We need to prove that the sum $d_{1}+d_{2}+\ldots+d_{n}$ does not depend on $P$: The area of the polygon $S=0.5... | proof | Geometry | proof | Yes | Yes | olympiads | false | 51,177 |
10.057. The diagonal of a rectangular trapezoid and its lateral side are equal. Find the length of the midline if the height of the trapezoid is 2 cm and the lateral side is $4 \mathrm{cm}$. | Solution.
Given $A D=2$ cm; $D B=B C=4$ cm (Fig. 10.56). In $\triangle D A B$, $\angle D A B=90^{\circ}$, $A B=\sqrt{16-4}=\sqrt{12}$, $B E=\frac{B K}{2}=\frac{A D}{2}=\frac{2}{2}=1$, $B M=\frac{B C}{2}=2$. In $\triangle B E M$, $\angle B E M=90^{\circ}$, $E M=\sqrt{4-1}=\sqrt{3}$, and $N M=N E+E M=A B+E M=$ $=\sqrt{1... | 3\sqrt{3}\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,178 |
10.058. A square with side length $m$ is inscribed in an equilateral triangle. Find the side length of the triangle. | Solution.
Given $E F=F N=N M=E M=m$ (Fig. 10.57); $A B=B C=A C=x$. $B O=\sqrt{x^{2}-\frac{x^{2}}{4}}=\frac{x \sqrt{3}}{2} \cdot \Delta B O C \sim \triangle F N C \Rightarrow \frac{x \sqrt{3}}{2 m}=\frac{2 x}{2(x-m)}, x=\frac{m(2 \sqrt{3}+3)}{3}$.
Answer: $\frac{m(2 \sqrt{3}+3)}{3}$. | \frac{(2\sqrt{3}+3)}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,179 |
10.060. The radii of the inscribed and circumscribed circles of a right triangle are 2 and 5 cm, respectively (Fig. 10.59). Find the legs of the triangle. | ## Solution.
Given $r=2 \text{cm}, R=5$ cm. In $\triangle A C B, \angle A C B=90^{\circ}, A B=2 R=10$, $A K=A E, K B=M B, A C=A E+C E=A K+C E=A K+r, B C=M B+C M=$ $=K B+C M=r+K B, A C+C B=2 r+A B, A C+C B=2(r+R)$. Let $A C=x$, then $C B=2(r+R)-x, C B=14-x$. Since $A B^{2}=A C^{2}+$ $+B C^{2}$, so $100=x^{2}+196-28 x+x... | 6;8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,181 |
10.061. A perpendicular line drawn from the vertex of a parallelogram to its diagonal divides this diagonal into segments of length 6 and $15 \mathrm{~cm}$. The difference in the lengths of the sides of the parallelogram is 7 cm (Fig. 10.60). Find the lengths of the sides of the parallelogram and its diagonals. | ## Solution.
By the condition $A E=6 \mathrm{~cm}, E C=15 \mathrm{~cm}, B C-A B=7$ cm. Let $B C=x$, $A B=x-7 \cdot \mathrm{B} \triangle B E A \angle B E A=90^{\circ} \Rightarrow B E^{2}=x^{2}-14 x+49-36$. In $\triangle B E C$ $\angle B E C=90^{\circ} \Rightarrow B E^{2}=x^{2}-225$. Therefore, $x^{2}-14 x+49-36=x^{2}-2... | 10;17;21;\sqrt{337}\mathrm{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,182 |
10.064. The lateral side of an isosceles triangle is $10 \mathrm{~cm}$, and the base is $12 \mathrm{~cm}$. Tangents are drawn to the inscribed circle of the triangle, parallel to the height of the triangle, and cutting off two right triangles from the given triangle. Find the lengths of the sides of these triangles. | ## Solution.
The area of the triangle is found using the formula (Fig. 10.63):
$$
S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{16(16-10)(16-10)(16-12)}=48
$$
The radius of the inscribed circle $r=\frac{S}{p}=\frac{48}{16}=3$ (cm); $r=D H . H C=D C-$ $-D H=6-3=3 \text{ cm} ; \triangle B C D$ and $\triangle F C H$ are similar, so ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,184 |
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