China Mathematical Competition

As seen in Fig. 5.1, connecting $PQ$, then $PQ$ is perpendicular to plane $ABC$ with the foot point $H$ being the center of $△ABC$. The center of the sphere $O$ is also on $PQ$. Connect and extend $CH$ to let it intersect with $AB$ at point $M$. $M$ is then the midpoint of $AB$, and $CM\perp AB$. It is easy to see that $\angle PMH$ and $\angle QMH$ are the plane angles formed by the sides-faces and the bases of the regular triangular pyramids $P$-$ABC$ and $Q$-$ABC$, respectively. Then $\angle PMH=45^{\circ }$, so $PH=MH=\frac{1}{2}AH$.



Since $\angle PAQ=90^{\circ }$, $AH\perp PQ$, then $A{H}^2=PH\cdot QH$. We then have $A{H}^2=\frac{1}{2}AH\cdot QH$. Therefore, $QH=2AH=4MH$. Finally, $\tan \angle QMH=\frac{QH}{MH}=4$.

The answer is 4.