13. [25] Four circles with radii $1,2,3$, and $r$ are externally tangent to one another. Compute $r$. (No proof is necessary.)
Answer: $6 / 23$
Solution: Let $A, B, C, P$ be the centers of the circles with radii $1,2,3$, and $r$, respectively. Then, $A B C$ is a 3-4-5 right triangle. Using the law of cosines in $\triangle P A B$ yields
$$
\cos \angle P A B=\frac{3^{2}+(1+r)^{2}-(2+r)^{2}}{2 \cdot 3 \cdot(1+r)}=\frac{3-r}{3(1+r)}
$$

Similarly,
$$
\cos \angle P A C=\frac{4^{2}+(1+r)^{2}-(3+r)^{2}}{2 \cdot 4 \cdot(1+r)}=\frac{2-r}{2(1+r)}
$$

We can now use the equation $(\cos \angle P A B)^{2}+(\cos \angle P A C)^{2}=1$, which yields $0=$ $23 r^{2}+132 r-36=(23 r-6)(r+6)$, or $r=6 / 23$.