China Mathematical Competition

As seen in Fig. 4.1, suppose the projections of the inscribed sphere's center $O$ on faces $ABC$ and $ABP$ are $H$, $K$, respectively, the midpoint of $AB$ is $M$, and the radius of the sphere is $r$. Then $P$, $K$, $M$ are collinear, $\angle PHM=\angle PKO=\frac{\pi }{2}$, and Fig. 4.1 $$\begin{gathered} OH=OK=r,PO=PH-OH=\sqrt{2}-r, \\ MH=\frac{\sqrt{3}}{6}AB=\frac{\sqrt{3}}{6},PM=\sqrt{M{H}^2+P{H}^2}=\sqrt{\frac{1}{12}+2}=\frac{5\sqrt{3}}{6}. \end{gathered}$$ Then we have $$\frac{r}{\sqrt{2}-r}=\frac{OK}{PO}=\sin \angle KPO=\frac{MH}{PM}=\frac{1}{5}.$$ Therefore, $r=\frac{\sqrt{2}}{6}$. The answer is $\frac{\sqrt{2}}{6}$.