 Touching Circles

Inside a right-angled triangle, two circles of the same radius are placed, each touching one of the legs, the hypotenuse, and the other circle. Find the radii of these circles if the legs of the triangle are equal to $a$ and $b$.

#
Let $\angle B A C=\alpha, \angle A B C=\beta$ be the acute angles of the given triangle. Express the hypotenuse in terms of the desired radius and trigonometric functions of the angles $\frac{\alpha}{2}$ and $\frac{\beta}{2}$.

## Solution

Let the circles with radii $r$ and centers $O_{1}$ and $O_{2}$ touch the hypotenuse $A B$ at points $M$ and $N$, respectively, with $B C=a, A C=b$, and the circle with center $O_{1}$ is inscribed in the angle $B A C$. Denote $A B=c, \angle B A C=\alpha, \angle A B C=\beta$. Then

$$
\begin{gathered}
\sin \alpha=\frac{a}{c}, \cos \alpha=\frac{b}{c}, \sin \beta=\frac{b}{c}, \cos \beta=\frac{a}{c} \\
\operatorname{tg} \frac{\alpha}{2}=\frac{\sin \alpha}{1+\cos \alpha}=\frac{a}{b+c}, \operatorname{tg} \frac{\beta}{2}=\frac{\sin \beta}{1+\cos \beta}=\frac{b}{a+c} \\
A M=\frac{O_{1} M}{\operatorname{tg} \frac{\alpha}{2}}=\frac{r(b+c)}{a}, B M=\frac{O_{2} N}{\operatorname{tg} \frac{\beta}{2}}=\frac{r(a+c)}{b}
\end{gathered}
$$

Since $c=A B=A M+M N+N B$ and $M N=O_{1} O_{2}=2 r$, we have the equation

$$
\frac{r(b+c)}{a}+2 r+\frac{r(a+c)}{b}=c
$$

from which we find that

$$
\begin{gathered}
r=\frac{c}{2+\frac{b+c}{a}+\frac{a+c}{b}}=\frac{a b c}{2 a b+b^{2}+b c+a^{2}+a c}=\frac{a b c}{(a+b)^{2}+c(a+b)}= \\
\frac{a b c}{(a+b)(a+b+c)}=\frac{a b \sqrt{a^{2}+b^{2}}}{(a+b)\left(a+b+\sqrt{a^{2}+b^{2}}\right)}
\end{gathered}
$$

## Answer

## Problem