8,9 Inscribed and Circumscribed Circles

In triangle $A B C$, side $B C$ is equal to $a$, and the radius of the inscribed circle is $r$.

Find the radii of two equal circles that touch each other, where one of them touches sides $B C$ and $B A$, and the other touches $B C$ and $C A$.
Let \( x \) be the desired radius.

First method. Let \( \angle B = 2 \beta, \angle C = 2 \gamma \). Then \( x \operatorname{ctg} \beta + x \operatorname{ctg} \gamma + 2 x = a \). Since \( r \operatorname{ctg} \beta + r \operatorname{ctg} \gamma = a \), we have \( \operatorname{ctg} \beta + \operatorname{ctg} \gamma = a / r \), and thus \( a x / r + 2 x = a \). Therefore,

\[ x = \frac{a r}{a + 2 r} \]

Second method. Let \( O_1 \) and \( O_2 \) be the centers of the given equal circles. Then the rays \( B O_1 \) and \( C O_2 \) intersect at the center \( O \) of the inscribed circle of triangle \( A B C \).

The ratio of the heights of similar triangles \( O O_1 O_2 \) and \( O B C \), drawn from the common vertex \( O \), is equal to the ratio of the corresponding sides, that is,

\[ \frac{r - x}{r} = \frac{2 x}{a} \]. Therefore, \( x = \frac{a r}{a + 2 r} \).

## Answer

\[ \frac{a r}{a + 2 r} \]