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 |
|---|---|---|---|---|---|---|---|---|---|
11.51 Find the range of the function $y=5 \sin x-12 \cos x$. | 11.51 In the coordinate plane $x O y$, let's construct the point $A(5 ; 12)$ and consider the vector $\overline{O A}$ (Fig. 11.18). Suppose this vector forms an angle $\varphi$ with the x-axis. Then, from the figure, it is clear that $\quad \cos \varphi=\frac{5}{\sqrt{25+144}}=\frac{5}{13}, \quad \sin \varphi=\frac{12}... | -13\leqy\leq13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,967 |
11.52 Which is greater: $\sin 2 x$ or $2 \sin x$? | 11.52 To compare the expressions $\sin 2 x$ and $2 \sin x$, we find their difference:
$A=\sin 2 x-2 \sin x=2 \sin x \cos x-2 \sin x=2 \sin x(\cos x-1)$.
Using the inequalities $-1 \leq \sin x \leq 1$ and $-1 \leq \cos x \leq 1$. Subtracting 1 from all parts of the second inequality, we get $-2 \leq \cos x-1 \leq 0$. ... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 49,968 |
11.53 What is the value of the product
$\log _{3} 2 \cdot \log _{4} 3 \cdot \log _{5} 4 \ldots \log _{10} 9$
if it is known that $\lg 2=0.3010$? | 11.53 Instruction. Convert all factors to base 10.
Answer: 0.3010. | 0.3010 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,969 |
11.54 It is known that $\log _{b} a=m$ and $\log _{c} b=n$. Find $\log _{b c}(a b)$.
保留源文本的换行和格式，直接输出翻译结果如下：
11.54 It is known that $\log _{b} a=m$ and $\log _{c} b=n$. Find $\log _{b c}(a b)$. | ### 11.54 We have:
\[
\begin{aligned}
& \log _{b c}(a b)=\log _{b c} a+\log _{b c} b \\
& \log _{b} a=\frac{\log _{b c} a}{\log _{b c} b} \Rightarrow \\
& \Rightarrow \log _{b c} a=m \log _{b c} b=m \frac{1}{1+\log _{b} c}=m \frac{1}{1+\frac{1}{n}}=\frac{m n}{n+1} \\
& \log _{c} b=\frac{\log _{b c} b}{\log _{b c} c} \... | \frac{n(+1)}{n+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,970 |
11.55 Simplify the fraction $\frac{n!}{(n+1)!-n!}$. | 11.55 Using the definition of factorial, we find
$$
\frac{n!}{(n+1)!-n!}=\frac{1 \cdot 2 \cdot 3 \cdot \ldots \cdot n}{n!(n+1-1)}=\frac{1 \cdot 2 \cdot 3 \cdot \ldots \cdot n}{n!n}=\frac{n!}{n!n}=\frac{1}{n}
$$
Answer: $\frac{1}{n}$. | \frac{1}{n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,971 |
12.1 $\lim _{x \rightarrow 2} \frac{x^{3}-8}{2 x-4}$. | 12.1 The function $f(x)=\frac{x^{3}-8}{2 x-4}$ is undefined at the point $x=2$. By factoring the numerator using formula (1.14), we can represent this function as
$f(x)=\frac{(x-2)\left(x^{2}+2 x+4\right)}{2(x-2)}$.
In the domain of the function $f(x)$, the expression $x-2 \neq 0$, so the fraction can be simplified b... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,972 |
$12.2 \lim _{x \rightarrow 4} \frac{\sqrt{x^{2}-7}-3}{x-4}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$12.2 \lim _{x \rightarrow 4} \frac{\sqrt{x^{2}-7}-3}{x-4}$. | 12.2 Instruction. Multiply the numerator and denominator of the fraction by $\sqrt{x^{2}-7}+3$
Answer: $\frac{4}{3}$. | \frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,973 |
$12.3 \lim _{x \rightarrow-1} \frac{\sqrt{2 x+3}-1}{\sqrt{5+x}-2}$.
$12.3 \lim _{x \rightarrow-1} \frac{\sqrt{2 x+3}-1}{\sqrt{5+x}-2}$.
The above text is already in a mathematical expression format, so the translation is the same as the original text. If you need an explanation or solution for the limit, please let ... | 12.3 Multiplying the numerator and the denominator of the fraction by $(\sqrt{2 x+3}+1)(\sqrt{5+x}+2)$, we get $\lim _{x \rightarrow-1} \frac{(2 x+3-1)(\sqrt{5+x}+2)}{(5+x-4)(\sqrt{2 x+3}+1)}=\lim _{x \rightarrow-1} \frac{2(x+1)(\sqrt{5+x}+2)}{(x+1)(\sqrt{2 x+3}+1)}=$ $\lim _{x \rightarrow-1} \frac{2(\sqrt{5+x}+2)}{\sq... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,974 |
12.4 Compute the limits, then verify or refute the statement
$$
\lim _{x \rightarrow 3} \frac{x^{2}+x-12}{3 x-9}=5+\lim _{x \rightarrow 2} \frac{x^{4}-16}{8-x^{3}}
$$
Find the derivatives of the functions (12.5-12.7):
| 12.4 We find
$$
\lim _{x \rightarrow 3} \frac{x^{2}+x-12}{3 x-9}=2 \frac{1}{3}(1) ; 5+\lim _{x \rightarrow 2} \frac{x^{4}-16}{8-x^{3}}=5-2 \frac{2}{3}=2 \frac{1}{3}
$$
From (1) and (2) it follows that the given equality is correct. | 2\frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,975 |
$12.5 y=\sqrt[3]{x^{2}-1}\left(x^{4}-1\right)$.
The above text is translated into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the provided text is already in English and consists of a mathematical equation, no translation is needed. The... | 12.5 Let's write the given function as $y=\left(x^{2}-1\right)^{\frac{1}{3}}\left(x^{4}-1\right)$, from which, using the product differentiation rule (12.3) and the table of derivatives, we find
$$
\begin{aligned}
y^{\prime} & =\frac{1}{3}\left(x^{2}-1\right)^{-\frac{2}{3}} \cdot 2 x\left(x^{4}-1\right)+\left(x^{2}-1\... | \frac{2x(x^{2}-1)(7x^{2}+1)}{3\sqrt[3]{(x^{2}-1)^{2}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,976 |
12.7 $y=\frac{1-\cos 2 x}{1+\cos 2 x}$.
Calculate the values of the derivatives of the given functions at the specified values of the independent variable (12.8-12.9): | 12.7 Let's use the quotient rule of differentiation (12.4) and the table of derivatives:
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1-\cos 2 x}{1+\cos 2 x}\right)^{\prime}= \\
& =\frac{(1-\cos 2 x)^{\prime}(1+\cos 2 x)-(1-\cos 2 x)(1+\cos 2 x)^{\prime}}{(1+\cos 2 x)^{2}}= \\
& =\frac{-(-2 \sin 2 x)(1+\cos 2 x)-(1-\co... | \frac{4\sin2x}{(1+\cos2x)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,978 |
12.8 $f(x)=\sqrt{x^{2}+3}+\frac{2 x}{x+1} ; f^{\prime}(1)=?$ | 12.8 First, we find
$$
f^{\prime}(x)=\frac{x}{\sqrt{x^{2}+3}}+2 \frac{x+1-x}{(x+1)^{2}}=\frac{x}{\sqrt{x^{2}+3}}+\frac{2}{(x+1)^{2}}
$$
Now, setting $x=1$, we get $f^{\prime}(1)=\frac{1}{2}+\frac{2}{4}=1$.
Answer: 1. | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,979 |
12.9 $f(x)=\sin 4 x \cos 4 x ; f^{\prime}\left(\frac{\pi}{3}\right)=$ ? | 12.9 We have
$f^{\prime}(x)=4 \cos 4 x \cdot \cos 4 x+\sin 4 x(-4 \sin 4 x)=$ $=4\left(\cos ^{2} 4 x-\sin ^{2} 4 x\right)=4 \cos 8 x$.
At $x=\frac{\pi}{3}$ we find $f^{\prime}\left(\frac{\pi}{3}\right)=4 \cos \frac{8 \pi}{3}=4 \cos \frac{2 \pi}{3}=-2$.
Answer: -2. | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,980 |
12.11 $f(x)=\frac{(x-2)^{2}}{x^{2}}$. | 12.11 The domain of the function is the entire number line, except for the point $x=0$. Using formula (12.4) and the table of derivatives, we find
$$
f^{\prime}(x)=\frac{2(x-2) x^{2}-(x-2)^{2} 2 x}{x^{4}}=\frac{4(x-2)}{x^{3}}
$$
$f^{\prime}(x)=0$ only at $x=2$. We will construct a table:
| Interval | $(-\infty, 0)$ ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,982 |
12.12 $f(x)=-x(x-3)^{2}$. | 12.12 The function is defined for all $x$. We have
$$
\begin{aligned}
& f^{\prime}(x)=-(x-3)^{2}-x \cdot 2(x-3)=-(x-3)(x-3+2 x)= \\
& =3(x-1)(3-x) ; \\
& f^{\prime}(x)=0 \text { when } x=1 \text { and } x=3 . \text { Let's construct a table: }
\end{aligned}
$$
| Interval | $(-\infty, 1)$ | 1 | $(1,3)$ | 3 | $(3, \inf... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,983 | |
12.14 $y=x^{3}-3 x^{2}+3 x+2$ on $[-2,2]$. | 12.14 First, we find the values of $y$ at the endpoints of the interval: $y(-2)=-24, y(2)=4$. Now let's find the critical points within the interval $[-2,2]$. We have $y^{\prime}=3 x^{2}-6 x+3=3(x-1)^{2} ;$ the equation $y^{\prime}=0$ has a root $x=1$, and $y(1)=3$. By comparing the values of $y(-2), y(2)$, and $y(1)$,... | y_{\text{}}=-24,y_{\text{max}}=4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,985 |
12.15 $f(x)=2 \sin x+\sin 2 x$ on $\left[0, \frac{3 \pi}{2}\right]$. | 12.15 First, let's find the values of $f(x)$ at the ends of the given interval: $f(0)=0, f\left(\frac{3 \pi}{2}\right)=-2$, and then the critical points belonging to this interval. We have $f^{\prime}(x)=2 \cos x+2 \cos 2 x ; f^{\prime}(x)=0$, if $\cos x+\cos 2 x=0$, from which $2 \cos \frac{3 x}{2} \cos \frac{x}{2}=0$... | y_{\text{}}=-2,y_{\text{max}}=\frac{3\sqrt{3}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,986 |
12.16 Split the number 18 into two addends such that the sum of their squares is the smallest possible. | 12.16 Let $x$ be the first addend; then $18-x$ is the second addend. According to the condition, we get the function $S(x)=x^{2}+(18-x)^{2}$, where by the context of the problem $00$. Thus, $x=9$ is the minimum point of $S(x)$, and therefore, each addend is equal to 9.
Answer: 9 and 9. | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,987 |
12.17 A rectangular plot of land with an area of $294 \mathrm{~m}^{2}$ needs to be fenced and then divided into two equal parts by another fence. What should the linear dimensions of the plot be for the total length of the fence to be the smallest? | 12.17 Let $x$ and $y$ (Fig. 12.2) (linear dimensions of the plot in meters), then the area of the plot is $x y=294$, from which $y=\frac{294}{x}$. The length of the entire fence
Fig. 12.2 c... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,988 |
12.18 A pedestrian set out for a walk from point $A$ at a speed of $v$ km/h. After he had walked 6 km from $A$, a cyclist set out after him from $A$ at a speed 9 km/h greater than the pedestrian's speed. When the cyclist caught up with the pedestrian, they turned back and returned to $A$ together at a speed of 4 km/h. ... | 12.18 The time it takes for the cyclist to catch up with the pedestrian is $\frac{6}{9}=\frac{2}{3}$ (hours). Before the meeting, the pedestrian had been walking for $\frac{6}{v}+\frac{2}{3}$ (hours) and covered $v\left(\frac{6}{v}+\frac{2}{3}\right)$ (km). They covered this same distance on the way back at a constant ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,989 |
12.19 Given the function $y=x^{4}-6 x^{2}+1$. Find the maximum and minimum values of its derivative on the interval $[-1,3]$.
| 12.19 We have $y^{\prime}=4 x^{3}-12 x=f(x)$. We find the values of the function $f(x)$ at the ends of the interval $[-1,3]: f(-1)=8, f(3)=72$. Now let's find the critical points of the function $f(x)$ that belong to this interval: $f^{\prime}(x)=y^{\prime \prime}=$ $=12 x^{2}-12=12\left(x^{2}-1\right) ; f^{\prime}(x)=... | y_{\text{}}^{\}=-8,y_{\text{max}}^{\}=72 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,990 |
12.20 Form the equation of the tangent line to the graph of the function $f(x)=\frac{x^{2}+4}{x-2}$ at the point of its intersection with the y-axis. | 12.20 According to formula (12.6), the equation of the tangent is written in the form $y-y_{0}=f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)$, where $\left(x_{0} ; y_{0}\right)$ is the point of tangency. The abscissa $x_{0}$ of the point of intersection of the graph with the $O$ axis is equal to zero, and the ordina... | -x-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,991 |
12.21 At what angle to the $O x$ axis is the tangent to the graph of the function $g(x)=x^{2} \ln x$ at the point $x_{0}=1$ inclined? | 12.21 The slope coefficient of the tangent to the graph of the function $g(x)$ is equal to $k=g^{\prime}\left(x_{0}\right)=\operatorname{tg} \alpha$, where $x_{0}$ is the abscissa of the point of tangency, and $\alpha$ is the angle of inclination of the tangent to the $O x$ axis. We find
$$
g^{\prime}(x)=2 x \ln x+x^{... | \frac{\pi}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,992 |
12.22 On the graph of the function $y=x(x-4)^{3}$, find the points where the tangents are parallel to the x-axis.
Find the points of extremum of the functions (12.23-12.24): | 12.22 Let $\left(x_{0} ; y_{0}\right)$ be the point of tangency. Since the tangent is parallel to the x-axis, the slope of the tangent at this point is zero. We have
$$
\begin{aligned}
& y^{\prime}=(x-4)^{3}+x \cdot 3(x-4)^{2}=(x-4)^{2}(x-4+3 x)= \\
& =4(x-4)^{2}(x-1) ; k=y^{\prime}\left(x_{0}\right)=4\left(x_{0}-4\ri... | (4;0)(1;-27) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,993 |
$12.23 y=\frac{x}{\ln x}$.
The above text is translated into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the provided text is already in a mathematical format which is universal and does not require translation, the translation is as f... | 12.23 The function is defined for $x>0$ and $x \neq 1$. We have
$$
y^{\prime}=\frac{\ln x - x \cdot \frac{1}{x}}{\ln ^{2} x}=\frac{\ln x - 1}{\ln ^{2} x}
$$
Thus, $y^{\prime}=0$ when $\ln x - 1 = 0$, i.e., when $x=e$. Let's construct a table:
| Interval | $(0,1)$ | $(1, e)$ | $e$ | $(e, \infty)$ |
| :---: | :---: | ... | (e) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,994 |
$12.24 y=x^{3} e^{-x}$. | 12.24 The function is defined for all $x$. We find
$$
y^{\prime}=3 x^{2} e^{-x}-x^{3} e^{-x}=x^{2} e^{-x}(3-x)
$$
The equation $y^{\prime}=0$ has only one root $x=3$. Since $y^{\prime}>0$ for $x<3$ and $y^{\prime}<0$ for $x>3$, the function reaches a maximum at the point $x=3$, which is equal to $3^{3} e^{-3}=\frac{2... | (3;\frac{27}{e^{3}}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,995 |
12.25 For the function $f(x)=2 \sin 5 x+\sqrt{x}+\frac{3}{5} \quad$ find the antiderivative $F(x)$ under the condition that the graphs of the functions $f(x)$ and $F(x)$ intersect at a point lying on the $O y$ axis. | 12.25 Since for the function $\sin x$ one of the antiderivatives is $-\cos x$, then, according to the rules of point $3^{0}$, the antiderivative of the function $2 \sin 5 x$ is $-\frac{2}{5} \cos 5 x$. Further, the antiderivative of the function $\sqrt{x}+\frac{3}{5}$ is $\frac{2}{3} x \sqrt{x}+\frac{3}{5} x$. Then for... | F(x)=-\frac{2}{5}\cos5x+\frac{2}{3}x\sqrt{x}+\frac{3}{5}x+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,996 |
12.26 Find the function $F(x)$, the graph of which passes through the point $M_{0}(3 ;-2)$, given that $F^{\prime}(x)=4 x^{2}+9 x^{-2}$.
Calculate the integrals (12.27-12.30): | 12.26 We have $F(x)=4 \frac{x^{3}}{3}+9 \frac{x^{-1}}{-1}+C$. Substituting the coordinates of the point $M_{0}(3 ;-2)$ into this equation, we get $-2=\frac{4}{3} 3^{3}-9 \cdot 3^{-1}+C$, from which $C=-35$.
Answer: $F(x)=\frac{4}{3} x^{3}-\frac{9}{x}-35$. | F(x)=\frac{4}{3}x^{3}-\frac{9}{x}-35 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,997 |
$12.27 \int_{0}^{\pi} \cos ^{2} x d x$ | 12.27 To transform the integrand, we will use the power reduction formula (4.17). Then we get
$$
\begin{aligned}
& \int_{0}^{\pi} \cos ^{2} x d x=\int_{0}^{\pi} \frac{1+\cos 2 x}{2} d x=\frac{1}{2}\left(\int_{0}^{\pi} d x+\int_{0}^{\pi} \cos 2 x d x\right)= \\
& =\frac{1}{2}\left(\left.x\right|_{0} ^{\pi}+\left.\frac{... | \frac{\pi}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,998 |
$12.28 \int_{8}^{27} \frac{d x}{\sqrt[3]{x^{2}}}$ | 12.28 We have
$$
\int_{8}^{27} \frac{d x}{\sqrt[3]{x^{2}}}=\int_{8}^{27} x^{-2 / 3} d x=\left.\frac{x^{1 / 3}}{1 / 3}\right|_{8} ^{27}=3\left(27^{1 / 3}-8^{1 / 3}\right)=3
$$
Answer: 3. | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 49,999 |
12.35 $\lim _{x \rightarrow 4} \frac{x+\sqrt{x}-6}{x-5 \sqrt{x}+6}$. | 12.35 The numerator and denominator of the given fraction are quadratic functions of $\sqrt{x}$. We have
$$
\lim _{x \rightarrow 4} \frac{(\sqrt{x}+3)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\lim _{x \rightarrow 4} \frac{\sqrt{x}+3}{\sqrt{x}-3}=\frac{5}{-1}=-5
$$
Answer: -5. | -5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,005 |
$12.36 \lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt[3]{x^{2}+2 \sqrt[3]{x}-3}}$.
Calculate the limits, and then confirm or refute the following statements ( $12.37-12.38$ ): | 12.36 Factor the denominator of the fraction:
$$
\sqrt[3]{x^{2}}+2 \sqrt[3]{x}-3=(\sqrt[3]{x}-1)^{2}+4 \sqrt[3]{x}-4=(\sqrt[3]{x}-1)(\sqrt[3]{x}+3)
$$
Now, by canceling the fraction by $\sqrt[3]{x}-1 \neq 0$, we find
$$
\lim _{x \rightarrow 1} \frac{1}{\sqrt[3]{x}+3}=\frac{1}{4}
$$
Answer: 0.25 . | 0.25 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,006 |
$12.39 f(x)=\frac{\cos x}{1+\sin x} ; f^{\prime}\left(\frac{\pi}{2}\right)=$ ? | 12.39 We have
\[
\begin{aligned}
& f^{\prime}(x)=\frac{-\sin x(1+\sin x)-\cos x \cdot \cos x}{(1+\sin x)^{2}}= \\
& =\frac{-\sin x-\sin ^{2} x-\cos ^{2} x}{(1+\sin x)^{2}}=-\frac{1+\sin x}{(1+\sin x)^{2}}=-\frac{1}{1+\sin x} \\
& f^{\prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2}
\end{aligned}
\]
Answer: $-0.5$. | -0.5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,008 |
$12.40 f(x)=\frac{2^{2 x}}{\sqrt{2-2^{2 x}}} ; f^{\prime}(0)=?$ | 12.40 Since $2^{2 x}=4^{x}$, then $f(x)=\frac{4^{x}}{\sqrt{2-4^{x}}}$. We find
$$
\begin{aligned}
& f^{\prime}(x)=\frac{4^{x} \ln 4 \cdot \sqrt{2-4^{x}}-\frac{4^{x}\left(-4^{x}\right) \ln 4}{2 \sqrt{2-4^{x}}}}{2-4^{x}}= \\
& =\frac{4^{x} \ln 4\left(4-2 \cdot 4^{x}+4^{x}\right)}{2\left(2-4^{x}\right) \sqrt{2-4^{x}}}=\f... | 3\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,009 |
12.41 Find the intervals of increase and decrease of the function $y=\sqrt{3} \sin x-\cos x$.
Find the minimum and maximum values of the functions on the given intervals (12.42-12.43): | 12.41 We have
\[
\begin{aligned}
& y^{\prime}=\sqrt{3} \cos x+\sin x=2\left(\sin \frac{\pi}{3} \cos x+\cos \frac{\pi}{3} \sin x\right)= \\
& =2 \sin \left(\frac{\pi}{3}+x\right)
\end{aligned}
\]
From the equation \(\sin \left(\frac{\pi}{3}+x\right)=0\) we find \(x=-\frac{\pi}{3}+\pi n\), \(n \in \mathbb{Z}\). Next, w... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,010 |
12.42 $f(x)=\frac{\sin 2 x}{\sin \left(\frac{\pi}{4}+x\right)}$ on $\left[0, \frac{\pi}{2}\right]$. | 12.42 First, we compute the values of $f(x)$ at the ends of the interval:
$$
\begin{aligned}
& f(0)=0, f\left(\frac{\pi}{2}\right)=0 . \text { Next, we find } \\
& f^{\prime}(x)=\frac{2 \cos 2 x \sin \left(\frac{\pi}{4}+x\right)-\sin 2 x \cos \left(\frac{\pi}{4}+x\right)}{\sin ^{2}\left(\frac{\pi}{4}+x\right)}
\end{al... | y_{\text{}}=0,y_{\text{max}}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,011 |
12.43 $f(x)=2^{\sqrt[3]{x^{2}}}:$ а) on $[-8,-1]$; б) on $[-1,1]$. | 12.43 Find \( f^{\prime}(x) = 2^{\sqrt[3]{x^{2}}} \ln 2 \cdot \frac{2}{3} x^{-1 / 3} \); \( f^{\prime}(x) \neq 0 \) for any \( x \), \( f^{\prime}(x) \) does not exist at \( x = 0 \).
a) If \( x \in [-8, -1] \), then \( f(-8) = 2^{4} = 16 \), \( f(-1) = 2 \);
thus, \( y_{\text{min}} = 2 \), \( y_{\text{max}} = 16 \).
... | )y_{}=2,y_{max}=16;b)y_{}=1,y_{max}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,012 |
12.44 In an arithmetic progression, the sixth term is equal to 3, and the difference of the progression is greater than 0.5. For what value of the difference of this progression is the product of the first, fourth, and fifth terms the greatest? | 12.44 According to the condition, $a_{6}=a_{1}+5 d=3$, from which $a_{1}=3-5 d$. Let the product $a_{1} a_{4} a_{5}$ be denoted by $y$. Then we get $y=a_{1}\left(a_{1}+3 d\right)\left(a_{1}+4 d\right)=-10 d^{3}+51 d^{2}-72 d+27$.
To find the value of $d$ at which the function $y$ takes its maximum value, we first find... | 2.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,013 |
12.45 In a right triangle with a hypotenuse of 24 cm and an angle of $60^{\circ}$, a rectangle is inscribed, with its base lying on the hypotenuse. What should be the lengths of the sides of the rectangle so that its area is maximized? | 12.45 According to the condition, $AB=24 \text{ cm}, \angle A=60^{\circ}$, from which $\angle B=30^{\circ}$ (Fig. 12.7). Let $MN=x$ and $LM=y$ be the linear dimensions of the rectangle $MNKL$ (in centimeters). Express $x$ in terms of $y$. From $\triangle KBL: \quad BL=KL \operatorname{ctg} 30^{\circ}=x \sqrt{3}$; from ... | 3\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,014 |
12.46 The magnitude of the angle at the base of an isosceles triangle is equal to $\alpha$. For what value of $\alpha$ is the ratio of the lengths of the radii of the inscribed and circumscribed circles the greatest? What is this ratio? | 12.46 Instruction. Let in triangle $ABC$ points $O$ and $O_{1}$ be the centers of the circumscribed and inscribed circles, and $R$ and $r$ the radii of these circles (Fig. 12.8). Express the ratio $\frac{r}{R}$ in terms of $\alpha$ and investigate the function $\frac{r}{R}=f(\alpha)=\sin 2 \alpha \tan \frac{\alpha}{2}$... | \alpha=\frac{\pi}{3};\frac{r}{R}=\frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,015 |
12.47 The lateral face of a regular quadrilateral pyramid has a constant given area and is inclined to the base plane at an angle $\alpha$. For what value of $\alpha$ is the volume of the pyramid the greatest? | 12.47 The volume of the pyramid \(S_{ABCD}\) (Fig. 12.9) is expressed by the formula \(V = \frac{1}{3} S_{ABCD} \cdot SO\). Let \(AD = a\), then \(S_{ABCD} = a^2\), and from \(\triangle SOK\) we find \(SO = OK \operatorname{tg} \angle SKO = \frac{a}{2} \operatorname{tg} \alpha\).
$ to the point $(x ; y)$ lying on the hyperbola is $\sqrt{x^{2}+y^{2}}$. Therefore, we nee... | 29.28;(5.4;5.4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,017 |
12.49 At what points are the tangents to the curve $y=\frac{x^{3}}{3}-x^{2}-x+1$ parallel to the line $y=2 x-1$? | 12.49 Let $\left(x_{0} ; y_{0}\right)$ be the point of tangency. The slope of the tangent line at this point is $k=2$. We find $y^{\prime}=x^{2}-2 x-1 ; \quad k=y^{\prime}\left(x_{0}\right)=x_{0}^{2}-2 x_{0}-1=2 . \quad$ Solving the equation $x_{0}^{2}-2 x_{0}-3=0$, we get $\left(x_{0}\right)_{1}=3,\left(x_{0}\right)_{... | (3;-2)(-1;\frac{2}{3}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,018 |
12.50 Find the equations of the tangents to the curves $y=2 x^{2}-5$ and $y=x^{2}-3 x+5$, drawn through the points of intersection of these curves. | 12.50 To find the points of intersection of the given curves, we solve the system of equations
$$
\left\{\begin{array}{l}
y=2 x^{2}-5 \\
y=x^{2}-3 x+5
\end{array}\right.
$$
from which we obtain $x_{1}=-5, x_{2}=2$ and $y_{1}=45, y_{2}=3$. Thus, we have two points of intersection: $A(-5 ; 45)$ and $B(2 ; 3)$. Next, we... | -20x-55,-13x-20,8x-13,x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,019 |
12.51 Given the function $y=\frac{1}{3} x^{3}-\frac{1}{2} x^{2}-2 x+3$. Find the intervals of its increase and decrease, extrema, and sketch its graph. | ### 12.51 Finding
$$
y^{\prime}=x^{2}-x-2=0=(x+1)(x-2)
$$
The equation $y^{\prime}=0$ has roots $x_{1}=-1$ and $x_{2}=2$. If $-\infty < x < -1$, then $y^{\prime} > 0$; if $-1 < x < 2$, then $y^{\prime} < 0$; if $x > 2$, then $y^{\prime} > 0$. Thus, at $x=-1$, we have $y_{\max }=\frac{25}{6}$, and at $x=2$, we have $y... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,020 | |
12.52 A point moves in a straight line according to the law $s(t)=\sqrt[3]{t^{2}}$. Show that its acceleration is inversely proportional to the square of the distance traveled. | 12.52 First, we find the velocity of the point $v=s^{\prime}(t)=\frac{2}{3} t^{-1 / 3}$, and then its acceleration
$$
a=v^{\prime}(t)=s^{\prime \prime}(t)=-\frac{2}{9 \sqrt[3]{t^{4}}}=-\frac{2}{9(s(t))^{2}}
$$
Thus, $a=\frac{k}{(s(t))^{2}}$, where $k=-\frac{2}{9}$. | \frac{-\frac{2}{9}}{(())^{2}} | Calculus | proof | Yes | Yes | olympiads | false | 50,021 |
12.53 Given the function $f(x)=|x|$. Write the expression for the antiderivative of the function. | 12.53 If $x \geq 0$, then $f(x)=x$ and $F(x)=0.5 x^{2}+C$; if $x<0$, then
$$
f(x)=-x \text { and } F(x)=-0.5 x^{2}+C
$$ | F(x)=\begin{cases}0.5x^2+C& | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,022 |
12.54 Find the function $S(x)$, if its derivative $S^{\prime}(x)=\frac{2}{\sqrt{5-x}}$ and $S(1)=-1$.
Calculate the integrals (12.55-12.58): | 12.54 Let's write the derivative in the form \( S^{\prime}(x)=2(5-x)^{-1 / 2} \). Then, using rule \( 3^{0} \) v), we find
\[
S(x)=2 \frac{1}{-1} \frac{(5-x)^{1 / 2}}{\frac{1}{2}}+C=-4 \sqrt{5-x}+C
\]
Setting \( x=1 \), we get \( S(1)=-4 \cdot 2+C=-1 \), from which \( C=7 \). Therefore, \( S(x)=7-4 \sqrt{5-x} \). | S(x)=7-4\sqrt{5-x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,023 |
$12.55 \int_{0}^{\pi / 4}(\sin 2 t-\cos 2 t)^{2} d t$. | 12.55 Raising the expression $\sin 2 t-\cos 2 t$ to the square and using known trigonometric formulas, we find
$$
\begin{aligned}
& \int_{0}^{\pi / 4}(\sin 2 t-\cos 2 t)^{2} d t= \\
& =\int_{0}^{\pi / 4}\left(\sin ^{2} 2 t-2 \sin 2 t \cos 2 t+\cos ^{2} 2 t\right) d t= \\
& =\int_{0}^{\pi / 4} d t-\int_{0}^{\pi / 4} \s... | 0.25(\pi-2) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,024 |
$12.56 \int_{0}^{7 / 3} \frac{x+1}{\sqrt[3]{3 x+1}} d x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$12.56 \int_{0}^{7 / 3} \frac{x+1}{\sqrt[3]{3 x+1}} d x$. | 12.56 We have
\[
\begin{aligned}
& \int_{0}^{7 / 3} \frac{x+1}{\sqrt[3]{3 x+1}} d x=\int_{0}^{7 / 3} \frac{3(x+1)}{3 \sqrt[3]{3 x+1}} d x=\frac{1}{3} \int_{0}^{7 / 3} \frac{3 x+1}{\sqrt[3]{3 x+1}} d x+ \\
& +\frac{1}{3} \int_{0}^{7 / 3} \frac{2}{\sqrt[3]{3 x+1}} d x=\frac{1}{3}\left(\int_{0}^{7 / 3}(3 x+1)^{2 / 3} d x... | \frac{46}{15} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,025 |
12.59 Find the area of the figure bounded by the lines $y=x^{3}-4 x$ and $y=0$ for $x \geq 0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 12.59 First, let's construct the graph of the function $y=x^{3}-4 x$ for $x \geq 0$. We find the extrema:
$y^{\prime}=3 x^{2}-4$;
$3 x^{2}-4=0 ;$
$x= \pm \frac{2}{\sqrt{3}} \approx \pm 1.1$.
For
=6$. We find the ordinate of the point of tangency: $y_{0}=2 \cdot 2^{2}-2 \cdot 2+1=5$.
Therefore, the equation of the tangent line is
$y-5=6(x-2)$
or
$y=6 x-7$.
The figure w... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 50,029 |
1.1. The lengths of the sides of a right-angled triangle form an arithmetic progression with a common difference of 1 cm. Find the length of the hypotenuse. | 1.1. Let $c$ be the length of the hypotenuse. Then the lengths of the legs are $c-1$ and $c-2$. We have
$$
(c-1)^{2}+(c-2)^{2}=c^{2}, \text { or } c^{2}-6 c+5=0
$$
from which $c=5$ (cm) (the second root of the equation does not satisfy the condition).
Answer: 5 cm. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,030 |
1.2. A point on the hypotenuse, equidistant from both legs, divides the hypotenuse into segments of 30 and 40 cm. Find the legs of the triangle. | 1.2. By the condition, $\angle C=90^{\circ}, A D=30 \text{ cm}$, $D B=40$ cm (Fig. 1.1). Let $A C=x, B C=y$. Since the point equidistant from the sides of the angle lies on its bisector, then $\frac{x}{y}=\frac{30}{40}$, i.e., $y=\frac{4 x}{3}$. But $x^{2}+y^{2}=A B^{2}$ and, therefore,
; the point of intersection of the medians divides each of them in the ratio $2: 1$, hence $A O=\frac{10}{3} \quad$ (cm). From $\triangle A O D$ we find $O D=\sqrt{A O^{2}-A D^{2}}=$ $=\sqrt{\left(\frac{10}{3}\right)^{2}-(2 \sqrt{2})... | 6\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,032 |
1.4. In a right-angled triangle, the bisector of an acute angle divides the opposite leg into segments of 4 and 5 cm. Determine the area of the triangle. | 1.4. Let $B C=x$ (Fig. 1.3). Then $A B=\sqrt{81+x^{2}}$ and $\frac{x}{4}=\frac{\sqrt{81+x^{2}}}{5}$ (since $B D$ is the bisector). From this, we have $25 x^{2}=$ $=16\left(81+x^{2}\right)$, i.e., $x=12$ (cm). Therefore,
$S_{\triangle A B C}=\frac{1}{2} B C \cdot A C=\frac{1}{2} \cdot 12 \cdot 9=54\left(\mathrm{~cm}^{2... | 54\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,033 |
1.5. The base of the triangle is $30 \mathrm{~cm}$, and the lateral sides are 26 and $28 \mathrm{~cm}$. The height is divided in the ratio $2: 3$ (measured from the vertex), and a line parallel to the base is drawn through the point of division. Determine the area of the resulting trapezoid. | 1.5. Given $A B=c=30 \mathrm{~cm}, A C=b=26 \mathrm{~cm}, B C=a=28 \mathrm{~cm}$ (Fig. 1.4). Then $p=0.5(a+b+c)=42, p-a=14, p-b=16, p-c=12$ and using Heron's formula, we find
$$
S_{\triangle A B C}=\sqrt{42 \cdot 14 \cdot 16 \cdot 12}=336\left(\mathrm{~cm}^{2}\right)
$$
Since $\triangle M N C \sim \triangle A B C$, t... | 282.24\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,034 |
1.6. The legs of a right triangle are 9 and 12 cm. Find the distance between the point of intersection of its angle bisectors and the point of intersection of the medians. | 1.6. In $\triangle A B C$ (Fig. 1.5), we have $A B=\sqrt{9^{2}+12^{2}}=15$ (cm); the median $C D$ is half the hypotenuse, i.e., $\frac{15}{2}$ cm. Let $E$ be the point of intersection of the medians; then $C E=\frac{2}{3} C D=\frac{2}{3} \cdot \frac{15}{2}=5$ (cm) (according to the additional relations, point $1^{\circ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,035 |
1.7. The area of triangle $ABC$ is $30 \, \text{cm}^2$. A point $D$ is taken on side $AC$ such that $AD:DC = 2:3$. The length of the perpendicular $DE$, drawn to side $BC$, is $9 \, \text{cm}$. Find $BC$. | 1.7. Draw $B D$ (Fig. 1.6); triangles $A B D$ and $B D C$ have a common height; therefore, their areas are in the ratio of the lengths of their bases, i.e., $S_{\triangle A B D}: S_{\triangle B D C}=2: 3$; then $S_{\triangle B D C}=\frac{3}{5} S_{\triangle A B C}=18 \mathrm{~cm}^{2} . \mathrm{C}$
On the other hand, ac... | 4\, | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,036 |
1.8. In an isosceles triangle, the heights drawn to the base and to the lateral side are equal to 10 and 12 cm, respectively. Find the length of the base. | 1.8. In $\triangle A B C$ we have $A B=B C$, $A E \perp B C, B D=10$ cm and $A E=12$ cm (Fig. 1.7). Let $A C=x, A B=B C=y$. Right triangles $A E C$ and $B D C$ are similar (angle $C$ is common); therefore, $B C: A C=B D: A E$, or $y: x=10: 12=5: 6$. Applying the Pythagorean theorem (1.13) to $\triangle B D C$, we have ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,037 |
1.9. Two circles touch each other externally. A tangent is drawn to the first circle, passing through the center of the second circle. The distance from the point of tangency to the center of the second circle is three times the radius of this circle. How many times greater is the length of the first circle compared to... | 1.9. Let $O_{1}$ and $O_{2}$ be the centers of the circles, and $A$ be the point of tangency (Fig. 1.8). Then $O_{1} A = R_{1}, O_{1} O_{2} = R_{1} + R_{2}, O_{2} A = 3 R_{2}$ (by the condition). We need to find the ratio $2 \pi R_{1} : 2 \pi R_{2} = R_{1} : R_{2}$. In the right triangle $O_{1} A O_{2} \left(\angle A =... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,038 |
1.10. Find the area of an isosceles triangle if its base is equal to $a$, and the length of the height drawn to the base is equal to the length of the segment connecting the midpoints of the base and the lateral side. | 1.10. By the condition, $D K-$ is the midline of $\triangle A B C$ (Fig. 1.9). Since $B D=D K=\frac{1}{2} B C$, then $\angle C=30^{\circ}$ and $B C=2 B D$. In $\triangle B C D$ we have $C D^{2}=B C^{2}-B D^{2}$, or $\frac{a^{2}}{4}=4 B D^{2}-B D^{2}$, from which $B D=\frac{a}{2 \sqrt{3}}$. Therefore,
. Then $A B=2 A E+E F$, or $2 x+x \sqrt{2}=a$, from which $x=\frac{a}{2+\sqrt{2}}=\frac{a(2-\sqrt{2})}{2}$. Therefore, the desired area is
$$
S=S_{A B C D}-4 S_{\triangle A E N}=a^{2}-\frac{4 x^{2}}{2}=a^{2}-\frac{4 a^{2}(4-4 \sqrt{2}+2)}{8}=2 a^{2}(\sqrt{2}-1)
$$
Answer: $2 a^{2}(\sqrt{2... | 2^{2}(\sqrt{2}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,040 |
1.12. A circle of radius $R$ is divided by two concentric circles into three figures of equal area. Find the radii of these circles. | 1.12. D e f i n i t i o n. Use the equalities $\pi R_{1}^{2}=S$, $\pi R_{2}^{2}=2 S, \pi R^{2}=3 S$, where $R_{1}$ and $R_{2}$ are the radii of the smaller and the middle circles, and $S$ is the area of the smaller circle.
A n s w e r: $R_{1}=\frac{R \sqrt{3}}{3}, R_{2}=\frac{R \sqrt{6}}{3}$. | R_{1}=\frac{R\sqrt{3}}{3},R_{2}=\frac{R\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,041 |
1.13. Inside a circle with a radius of 15 cm, a point $M$ is taken at a distance of 13 cm from the center. A chord of length 18 cm is drawn through this point. Find the lengths of the segments into which point $M$ divides the chord. | 1.13. Draw $O C \perp A B$ (Fig.
1.11). Then $C B=\frac{1}{2} A B=9 \quad \text{cm}$.
From $\triangle O B C$ it follows that
$O C=\sqrt{O B^{2}-B C^{2}}=12(\text{cm})$, and from
$\triangle O M C$ - that $M C=\sqrt{O M^{2}-O C^{2}}=5$ (cm). Therefore, $A M=9+5=14$ (cm), $M B=9-5=4$ (cm).
, then the area enclosed between the three semicircles is equal to the area of a circle whose diameter is the length of the perpendicular from the point of di... | 1.14. Let $R$ be the radius of the given semicircle, and $r$ be the radius of one of the constructed semicircles (Fig. 1.12). Then the area of the given figure is
$$
\begin{gathered}
0.5\left(\pi R^{2}-\pi r^{2}-\pi(R-r)^{2}\right)= \\
=\pi r(R-r)
\end{gathered}
$$
is the difference between the areas of sector \( A O B \) and triangle \( A O B \) (Fig. 1.13). We find \( S_{\text{sect } A O B} = \frac{\pi R^2}{3} \), \( S_{\triangle A O B} = \frac{1}{2} a \cdot \frac{R}{2} = \frac{a R}{4} \), hence \( S = \frac{\pi R^2}{3} - \frac{a R}{4} ... | \frac{^2(4\pi-3\sqrt{3})}{36} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,044 |
1.16. 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. | 1.16. Triangle $ABC$ is equilateral (Fig. 1.14); therefore, its area is $\frac{a^{2} \sqrt{3}}{4}$, where $a$ is the side of the triangle. Draw $OK \perp AC$. Since $\angle OAK=30^{\circ}$, then $AO=2r$, $AK=r \sqrt{3}$, hence $a=2r \sqrt{3} + 2r = 2r(\sqrt{3} + 1)$. Therefore,
$$
S=\frac{4r^{2}(3+2\sqrt{3}+1)\sqrt{3}... | 2r^{2}(2\sqrt{3}+3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,045 |
1.17. One of the two parallel lines touches a circle of radius $R$ at point $A$, while the other intersects this circle at points $B$ and $C$. Express the area of triangle $A B C$ as a function of the distance $x$ between the lines. | 1.17. Let the distance between the parallel lines be $x$ (Fig. 1.15); then the area $S$ of triangle $A B C$ is $0.5 B C \cdot x$. Since $O A \perp B C$, then $B M=$ $=M C$ and $B M \cdot M C=A M \cdot M D=x(2 R-$ $-x)$. Therefore, $0.25 B C^{2}=x(2 R-x)$, from which
$$
\begin{gathered}
S=0.5 x \cdot 2 \sqrt{2 R x-x^{2... | x\sqrt{2Rx-x^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,046 |
1.18. The radius of the circle circumscribed around a right triangle is to the radius of the inscribed circle as 5:2. Find the area of the triangle if one of its legs is equal to $a$. | 1.18. Let \( r \) and \( R \) be the radii of the inscribed and circumscribed circles, \( BC = a \) (Fig. 1.16). Let \( BD = x \); then \( BL = x \) (as tangents drawn from the same point), \( LA = AK = 2R - x \) (since \( \triangle ABC \) is a right triangle, then \( AB = 2R \)). We have \( AC^2 + BC^2 = AB^2 \).
or
... | \frac{2a^2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,047 |
1.19. Find the ratio of the radius of the circle inscribed in an isosceles right triangle to the height dropped to the hypotenuse. | 1.19. Since $\triangle ABC$ is an isosceles right triangle, the height $CD$ is also the bisector, i.e., $\angle DCA = \angle A = 45^\circ$ (Fig. 1.17); therefore, $AD = DC$ and $AC = \sqrt{2} DC$. But $AC = AK + KC = DC + r$ (since $AK = AD$ as tangents drawn from the same point),
:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,048 |
1.20. 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. | 1.20. Let's find the length of the side $B C$ (Fig. 1.18); $B C=\sqrt{B M^{2}+M C^{2}}=\sqrt{6^{2}+8^{2}}=10$ (cm). Considering that $A O$ is the bisector of $\triangle A B M$, we have $\frac{M O}{O B}=\frac{A M}{A B}$, or $\frac{r}{8-r}=\frac{6}{10}$, from which $r=3$ (cm). Since $D E \| A C$, then $\triangle D B E \s... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,049 |
1.21. 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. | 1.21. According to the condition, $R=0.5 \sqrt{12}$. The side $A B$ of the inscribed regular triangle (Fig. 1.19) is equal to $R \sqrt{3}$, i.e., $0.5 \sqrt{12} \cdot \sqrt{3}=3$. Let's find the side $C D$ of the new triangle: $a=\sqrt{3^{2}-1.5^{2}}=1.5 \sqrt{3}$. Since the radius $r$ of the inscribed circle is $\frac... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,050 |
1.22. An equilateral triangle has an inscribed circle. Three small circles are tangent to this circle and the sides of the triangle. Find the side of the triangle if the radius of the small circle is $r$.
保留源文本的换行和格式，直接输出翻译结果。 | 1.22. Let $a$ be the side of a triangle, $R$ be the radius of the inscribed circle; then $R=\frac{a \sqrt{3}}{6}$.
Draw radii $O M$ and $O_{1} K$ to the points of tangency (Fig. 1.20). From the similarity of triangles $A O M$ and $A O_{1} K$,
we have $\frac{R}{r}=\frac{A O}{A O-R-r}$.
From this, considering that
Fig. 1.23 $B N \perp A D$ (Fig. 1.24); since $B N=2 O M$, then $B N=\sqrt{75}$ cm. Considering that in $\triangle A N B \angle A B N=30^{\circ}$, we have $A B=2 A N$... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,057 |
1.29. Through points $R$ and $E$, belonging to sides $A B$ and $A D$ of parallelogram $A B C D$ and such that $A R=\frac{2}{3} A B, A E=\frac{1}{3} A D$, a line is drawn. Find the ratio of the area of the parallelogram to the area of the resulting triangle. | 1.29. Let $h$ be the height of parallelogram $ABCD$, and $h_{1}$ be the height of triangle $ARE$ (Fig. 1.25). Then $S_{ABCD} = AD \cdot h$, $S_{\triangle ARE} = \frac{1}{2} AE \cdot h_{1}$. But $\frac{h}{h_{1}} = \frac{AB}{AR} = \frac{3}{2}$. Therefore,
. 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 $h=\sqrt{20^{2}-x^{2}}=\sqrt{13^{2}-y^{2}}$. Considering that $y=25-4-$ $-x=21-x$, we hav... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,059 |
1.31. The diagonal of an isosceles trapezoid bisects its obtuse angle. The smaller base of the trapezoid is $3 \mathrm{~cm}$, and the perimeter is $42 \mathrm{~cm}$. Find the area of the trapezoid. | 1.31. By the condition, $\angle B C A=$ $=\angle A C D$ (Fig. 1.27). But $\angle B C A=$ $=\angle C A D$, so $\triangle A C D$ is isosceles and $A D=C D$. We have $3 A D+B C=42$; since $B C=3$ cm, then $A D=13$ cm. Draw $B K \perp A D$; then $A K=\frac{1}{2}(13-3)=5$ (cm) and from $\triangle A K B$ we find $B K=\sqrt{1... | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,060 |
1.32. Find the diagonal and the lateral side of an isosceles trapezoid with bases 20 and $12 \mathrm{~cm}$, given that the center of the circumscribed circle lies on the larger base of the trapezoid. | 1.32. Since $A D$ is the diameter of the circle (Fig. 1.28), then $O D = O C = 10 \, \text{cm}$. Draw $C L \perp A D$; then $O L = 6$ cm and from $\triangle C L O$ we find $A$ $C L = \sqrt{O C^{2} - O L^{2}} = 8 \, (\text{cm})$. Now from $\triangle A L C$ and $\triangle C L D$ we get
, the inscribed angle $C A D$ is $30^{\circ}$. Therefore, $h=\frac{1}{2} A C$ and from
Fig. 1.29
$\triangle A K C$ we get ... | ^{2}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,062 |
1.35. Calculate the ratio of the areas of a square, an equilateral triangle, and a regular hexagon inscribed in the same circle.
## Group B | 1.35. Let $R$ be the radius of the circle. Then the side of the inscribed regular triangle is $a_{3}=R \sqrt{3}$ and $S_{3}=\frac{a_{3}^{2} \sqrt{3}}{4}=\frac{3 R^{2} \sqrt{3}}{4}$. Next, the side of the square is $a_{4}=R \sqrt{2}$ and $S_{4}=a_{4}^{2}=2 R^{2}$, and finally, the side of the inscribed regular hexagon i... | 8:3\sqrt{3}:6\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,064 |
1.36. Inside a right angle, there is a point $M$, the distances from which to the sides of the angle are 4 and $8 \mathrm{~cm}$. A line passing through point $M$ cuts off a triangle from the right angle with an area of $100 \mathrm{~cm}^{2}$. Find the legs of the triangle. | 1.36. According to the condition, $\angle C=90^{\circ}, M P=$ $=4 \text{ cm}, M Q=8 \text{ cm}, S_{\triangle A B C}=100 \text{ cm}^{2}$ (Fig. $1.30$); it is required to find $B C$ and $A C$. Let $B C=x, A C=y$; then $0.5 x y=100$, i.e., $x y=200$. Since $\triangle B P M \sim \triangle M Q A$, we have
$$
\frac{M P}{A Q... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,065 |
1.37. The perimeter of a right triangle \(ABC (\angle C = 90^\circ)\) is 72 cm, and the difference between the lengths of the median \(CM\) and the altitude \(CK\) is 7 cm. Find the length of the hypotenuse. | 1.37. According to the condition, $\angle C=90^{\circ}, A B+B C+A C=72 \text{ cm}, C M$ is the median, $C K$ is the altitude, $C M-C K=7$ cm (Fig. 1.31); we need to find $A B$. Since $M$ is the center of the circumscribed circle,
}{2}$ (the second root of the equation does not satisfy the condition). Next, we find $x^{2}=\frac{c^{2}(3-\sqrt{5})}{2}$. Denoting the height to the hypot... | 0.5^{2}\sqrt{\sqrt{5}-2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,067 |
1.39. The heights of a triangle are 12, 15, and 20 cm. Prove that the triangle is a right triangle. | 1.39. Let the area of the triangle be $S$. Then its sides are: $a=\frac{2 S}{15}, b=\frac{2 S}{20}, c=\frac{2 S}{12}$. We find
$$
a^{2}+b^{2}=4 S^{2}\left(\frac{1}{15^{2}}+\frac{1}{20^{2}}\right)=4 S^{2}\left(\frac{1}{225}+\frac{1}{400}\right)=\frac{4 S^{2}}{25}\left(\frac{1}{9}+\frac{1}{16}\right)=\frac{4 S^{2}}{144}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,068 |
1.40. Point $M$ lies inside an equilateral triangle $A B C$. Calculate the area of this triangle, given that $A M=$ $=B M=2 \text{ cm}, \text{a} C M=1 \text{cm}$. | 1.40. Let the side of the triangle be denoted by $a$ and draw $M D \perp A B$ (Fig. 1.32). Since $A M = B M$, the points $C, M$, and $D$ lie on the same line - the height $C D$. In $\triangle A C D$ and $\triangle A M D$, we have $(1 + M D)^{2} = a^{2} - \frac{a^{2}}{4}$, $M D^{2} = 4 - \frac{a^{2}}{4}$. Then $a^{2} = ... | \frac{9\sqrt{3}+3\sqrt{15}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,069 |
1.41. The ratio of the magnitudes of two angles of a triangle is 2, and the difference in the lengths of the opposite sides is 2 cm; the length of the third side of the triangle is 5 cm. Calculate the area of the triangle. | 1.41. Given that $\angle A=2 \angle B, B C-A C=2$ cm, $A B=5$ cm (Fig. 1.33); we need to find $S_{\triangle A B C}$. Draw the angle bisector $A D$; then $\triangle A B C \sim \triangle A D C(\angle C$ - common, $\angle B=\angle D A C)$ and therefore $\frac{A C}{B C}=\frac{C D}{A C}$, i.e.
$$
A C^{2}=B C \cdot C D
$$
... | 3.75\sqrt{7}\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,070 |
1.42. The sides of a triangle are 13, 14, and 15 cm. Determine the areas of the triangles into which the given triangle is divided by its medians. | 1.42. First, let's prove that all the triangles mentioned in the problem are equal in area. We have \( S_{\triangle A O M} = S_{\triangle C O M} \) (Fig. 1.34), since
Fig. 1.34
these trian... | 14\,^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,071 |
1.43. The point of tangency of the inscribed circle in a right triangle divides the hypotenuse into segments of length $\boldsymbol{m}$ and n. Prove that the area of the triangle $S=m n$. Find the area of the rectangle inscribed in the given triangle such that one of its vertices coincides with the vertex of the right ... | 1.43. Let $D, E, F$ be the points of tangency (Fig. 1.35); then $A D=A F=m, B D=B E=n, C E=C F=r$ is the radius of the inscribed circle, $p=$ $=r+m+n-$ is the semiperimeter. Further, using formula (1.9), we find $S=\frac{(r+m)(r+n)}{2}$, or
$2 S=r^{2}+r(m+n)+m n=r(r+m+n)+m n=r p+m n$.
Since by equality (1.4) $r p=S$,... | \frac{2^{2}n^{2}}{(+n)^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 50,072 |
1.44. In a circle with center $O$, a chord $A B$ is drawn, intersecting the diameter at point $M$ and forming an angle of $60^{\circ}$ with the diameter. Find $O M$, if $A M=10$ cm, and $B M=4$ cm. | 1.44. Draw $O P \perp A B$ (Fig. 1.36). Then $A P=B P=7$ cm and, therefore, $M P=3$ cm. Since $\angle P M O=60^{\circ}$, then $\angle M O P=30^{\circ}$ and $O M=2 M P=6 \text{ cm}$.
Answer: 6 cm. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,073 |
1.45. In a circle of radius $R$, two parallel chords are drawn on opposite sides of the center, one of which subtends an arc of $60^{\circ}$, the other an arc of $120^{\circ}$. Find the area of the part of the circle enclosed between the chords. | 1.45. The area of the segment with an arc of $60^{\circ}$ is $S_{1}=\frac{\pi R^{2}}{6}-\frac{R^{2} \sqrt{3}}{4}$, and the area of the segment with an arc of $120^{\circ}$ is
Fig. 1.36 $S_{... | 0.5R^{2}(\pi+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,074 |
1.46. Find the radius of the circle, in the segment of which, corresponding to a chord of length $6 \mathrm{~cm}$, a square with a side of $2 \mathrm{~cm}$ is inscribed. | 1.46. Let $r$ be the desired radius. In $\triangle A O K$ (Fig. 1.37) we have $O K=\sqrt{r^{2}-9}$, and in $\triangle O B N$ we have
Fig. 1.37
$O N^{2}+B N^{2}=O B^{2}$, or $(O K+2)^{2}+1=... | \sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,075 |
1.47. On the segment $A B$, a point $C$ is taken, and on the parts $A C$ and $C B$ of this segment, semicircles are constructed with these parts as diameters. Prove that the sum of the lengths of these semicircles does not depend on the position of point $C$ on segment $A B$. | 1.47. Let the length of the segment $AB$ be $2l$. Denote the radius of one of the circles by $x$; then the radius of the second circle is $l-x$. The sum of the lengths of the semicircles is $L=\pi x + \pi(l-x) = \pi l$, i.e., it does not depend on $x$.
, so the length of the common chord $\mathrm{O}_{1} \mathrm{O}_{2}=R$. The desired area is twice the area of the segment with a central angle of $60^{\circ}$, i.e.,
$$
S=2\left(\frac{\pi R^{2}}{6}-\frac{R^{2} \sqrt{3}}{4}\right)=... | \frac{R^{2}(2\pi-3\sqrt{3})}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,077 |
1.49. The area of a right triangle is $24 \mathrm{~cm}^{2}$, and the hypotenuse is 10 cm. Find the radius of the inscribed circle. | 1.49. I n d i c a t i o n. Denoting the legs of the triangle by $a$ and $b$, solve the system of equations
$$
\left\{\begin{array}{l}
a b=48 \\
a^{2}+b^{2}=100
\end{array}\right.
$$
and then use the formula $S=p r$.
Answer: 2 cm. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,078 |
1.50. The center of a semicircle inscribed in a right triangle, such that its diameter lies on the hypotenuse, divides the hypotenuse into segments of 30 and 40. Find the length of the arc of the semicircle enclosed between the points of tangency with the legs. | 1.50. Draw radii $O D$ and $O E$ to the points of tangency (Fig. 1.39). We have $O D=O E=C E=C D$, i.e., $E C D O$ is a square. Let $R$ be the radius of the circle; then the length of the arc $E D$ is $\frac{\pi R}{2}$. Since $\triangle A E O \sim \triangle O D B$, we have $\frac{A E}{O D}=\frac{A O}{O B}=\frac{30}{40}... | 12\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,079 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.