Solution:

Let $O$ be the center of the sphere that is tangent to the faces $ABC$, $ABD$, and $BCD$. Let $P, Q$ be the feet of the perpendiculars from $O$ to $ABC$ and $ABD$ respectively. Let $R$ be the foot of the perpendicular from $P$ to $AB$. Then, $OPRQ$ is a quadrilateral such that $\angle P$, $\angle Q$ are right angles and $OP = OQ = r$. Also, $\angle R$ is the dihedral angle between faces $ABC$ and $ABD$, so $\cos \angle R = 1/3$. We can then compute $QR = \sqrt{2} r$, so $BR = \sqrt{6} r$. Hence, $1 = AB = 2(\sqrt{6} r) + 2r = 2r(\sqrt{6} + 1)$, so $r = (\sqrt{6} - 1)/10$.