The hypotenuse of a right triangle is $c$, and one of the acute angles is $\boldsymbol{\alpha}$. Two circles of the same radius are placed in the triangle, each touching one of the legs, the hypotenuse, and the other circle. Find the radii of these circles.
Express the hypotenuse in terms of the desired radius and the given angle.

## Solution

Let the circles of radii $r$ with centers $O_{1}$ and $O_{2}$ touch the hypotenuse $A B$ at points $M$ and $N$ respectively, and $\angle B A C=\alpha$. If the circle with center $O_{1}$ is inscribed in the angle $B A C$, then $A M=O_{1} M \boldsymbol{c t g} \angle M A O_{1}=r \operatorname{ctg} \frac{\alpha}{2}$.

Similarly, we find that $B N=r \operatorname{ctg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$. Since $c=A B=A M+M N+N B$ and $M N=O_{1} O_{2}=2 r$, we have the equation

$$
r \operatorname{ctg} \frac{\alpha}{2}+2 r+r \operatorname{ctg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)=c
$$

from which we find that $r=\frac{c}{2+\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}$.

## Answer