9. Four spheres, each of radius $r$, lie inside a regular tetrahedron with side length 1 such that each sphere is tangent to three faces of the tetrahedron and to the other three spheres. Find $r$.
Answer: $\frac{\sqrt{6}-1}{10}$
Solution: Let $O$ be the center of the sphere that is tangent to the faces $A B C, A B D$, and $B C D$. Let $P, Q$ be the feet of the perpendiculars from $O$ to $A B C$ and $A B D$ respectively. Let $R$ be the foot of the perpendicular from $P$ to $A B$. Then, $O P R Q$ is a quadrilateral such that $\angle P, \angle Q$ are right angles and $O P=O Q=r$. Also, $\angle R$ is the dihedral angle between faces $A B C$ and $A B D$, so $\cos \angle R=1 / 3$. We can then compute $Q R=\sqrt{2} r$, so $B R=\sqrt{6} r$. Hence, $1=A B=2(\sqrt{6} r)+2 r=2 r(\sqrt{6}+1)$, so $r=(\sqrt{6}-1) / 10$.