HMMT November 2013 U28

Triangle $A B C$ has $A B=4, B C=3$, and a right angle at $B$. Circles $\omega_{1}$ and $\omega_{2}$ of equal radii are drawn such that $\omega_{1}$ is tangent to $A B$ and $A C, \omega_{2}$ is tangent to $B C$ and $A C$, and $\omega_{1}$ is tangent to $\omega_{2}$. Find the radius of $\omega_{1}$. of $\omega_{1}$ and $\omega_{2}$ respectively. Suppose $\omega_{i}$ hits $A C$ at $B_{i}$ for $i=1,2$, so that $O_{1} O_{2}=B_{1} B_{2}=2 r$.

Extend angle bisector $A O_{1}$ to hit $B C$ at $P$. By the angle bisector theorem and triangle similarity $\triangle A B_{1} O_{1} \sim \triangle A B P$, we deduce $\frac{r}{A B_{1}}=\frac{B P}{A B}=\frac{3}{4+5}$. Similarly, $\frac{r}{C B_{2}}=\frac{4}{3+5}$, so

$$5=A C=A B_{1}+B_{1} B_{2}+B_{2} C=3 r+2 r+2 r=7 r$$

or $r=\frac{5}{7}$.