We have a triangle ABC such that
$$\mathrm { AB } = 9 , \quad \mathrm { BC } = 12 , \quad \angle \mathrm { ABC } = 90 ^ { \circ } .$$
There are also two circles $\mathrm { O } _ { 1 }$ and $\mathrm { O } _ { 2 }$ with radii of length $2r$ and $r$, respectively. The two circles $\mathrm { O } _ { 1 }$ and $\mathrm { O } _ { 2 }$ are tangential to each other. Further, $\mathrm { O } _ { 1 }$ is tangent to the two sides AB and AC, and $\mathrm { O } _ { 2 }$ is tangent to the two sides CA and CB. We are to find the value of $r$.
First, let D and E denote the points at which the segment AC is tangent to the circles $\mathrm { O } _ { 1 }$ and $\mathrm { O } _ { 2 }$ respectively, and set $\alpha = \angle \mathrm { O } _ { 1 } \mathrm { AC }$. Then, since $\tan 2 \alpha = \frac { \square \mathbf { A } } { \square }$, we have $\tan \alpha = \frac { \mathbf { C } } { \mathbf { D } }$ using the double-angle formula. Thus, we obtain $\mathrm { AD } = \mathbf { E }$.
Next, set $\beta = \angle \mathrm { O } _ { 2 } \mathrm { CA }$. Since $\alpha + \beta = \mathbf { F G } ^ { \circ }$, we have $\tan \beta = \frac { \mathbf { H } } { \square \mathbf{I} }$ using the addition theorem. Thus, we obtain $\mathrm { CE } = \square r$.
Moreover, it follows that $\mathrm { AC } = \mathbf { K L }$ and $\mathrm { DE } = \mathbf { M } \sqrt { \mathbf { N } } r$.
Finally we obtain
$$r = \frac { \mathbf { O P } ( \mathbf { Q } - \mathbf { R } \sqrt { \mathbf { S } } ) } { 41 }$$