A circle is inscribed in a circular sector with a central angle of $120^{\circ}$. Find its radius if the radius of the given circle is 8.
The line connecting the centers of two tangent circles passes through the point of tangency.

## Solution

Let $x$ be the radius of the desired circle, $O_{1}$ its center, $O A$ and $O B$ the radii of the given sector, $M$ the point of tangency of the circles, and $K$ the point of tangency of the desired circle with the radius $O B$. Then

$$
\begin{gathered}
O O_{1}=R-x, K O_{1}=x, \angle K O O_{1}=60^{\circ} \\
K O_{1}=O O_{1} \sin \angle K O O_{1}, \text { or } x=\frac{(R-x) \sqrt{3}}{2}
\end{gathered}
$$

From this, we find that

$$
x=\frac{R \sqrt{3}}{2+\sqrt{3}}=R \sqrt{3}(2-\sqrt{3}) .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-19.jpg?height=377&width=366&top_left_y=2055&top_left_x=846)

Answer

$R \sqrt{3}(2-\sqrt{3})$.

The angle with vertex $A$ is $\alpha$. The distance between the feet of the perpendiculars dropped from some point $B$ to the sides of the angle is $a$. Find $A B$.

## Hint

The points $A, B$ and the feet of the specified perpendiculars lie on the same circle.

## Solution

The segment $A B$ is seen from the feet $M$ and $N$ of the specified perpendiculars at an angle of $90^{\circ}$, so the points $A, B, M$ and $N$ lie on a circle with diameter $A B$, and the segment $M N$ is seen from point $A$ at an angle $\alpha$. Therefore,

$$
A B=\frac{M N}{\sin \alpha}=\frac{a}{\sin \alpha}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-20.jpg?height=298&width=548&top_left_y=637&top_left_x=755)

## Answer

$\frac{a}{\sin \alpha}$.