Problem (Geometry):
In triangle $A B C, C$ is a right angle and $M$ is on $\overline{A C}$. A circle with radius $r$ is centered at $M$, is tangent to $\overline{A B}$, and is tangent to $\overline{B C}$ at $C$. If $A C=5$ and $B C=12$, compute $r$.

Solution:
Let $N$ be the point of tangency of the circle with $\overline{A B}$ and draw $\overline{M B}$, as shown below.



Because $\triangle B M C$ and $\triangle B M N$ are right triangles sharing a hypotenuse, and $\overline{M N}$ and $\overline{M C}$ are radii, $\triangle B M C \cong \triangle B M N$. Thus $B N=12$ and $A N=1$. Also $\triangle A N M \sim \triangle A C B$ because the right triangles share $\angle A$, so $\frac{N M}{A N}=\frac{C B}{A C}$. Therefore $\frac{r}{1}=\frac{12}{5}$, so $r=\frac{\mathbf{1 2}}{\mathbf{5}}$.

Let $r$ denote the radius of the circle, and let $D$ be the foot of the perpendicular from $O$ to $\overline{A B}$. Note that $\triangle A B C \sim \triangle A O D$. Thus $\frac{A B}{A O}=\frac{B C}{D O} \Longrightarrow \frac{13}{5-r}=\frac{12}{r}$, and $r=\frac{\mathbf{1 2}}{\mathbf{5}}$.

Final answer: $\frac{12}{5}$