12.224. In a regular triangular pyramid, the dihedral angle at the base is equal to $\alpha$, the lateral surface area is $S$. Find the distance from the center of the base to the lateral face.
Solution.

Let $L O$ be the height of the regular pyramid $L A B C$ (Fig. 12.91). Draw a perpendicular $L D$ to $A C$ in the face $A L C$. Then $O D \perp A C, \angle L D O=\alpha$, side $A C$ is perpendicular to the plane $L D O$, and planes $L D O$ and $A L C$ are perpendicular.

In the plane $L D O$, draw a perpendicular $O K$ to $L D$. Then $O K$ is perpendicular to the plane $A L C$ and its length is the distance from point $O$ to the face $A L C$. Since the pyramid is regular, all lateral faces are equally inclined to the base plane and $S_{\triangle A B C}=S \cos \alpha$.

If $A C=a$, then $S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}, \frac{a^{2} \sqrt{3}}{4}=S \cos \alpha, a=2 \sqrt{\frac{S \cos \alpha}{\sqrt{3}}}$.

$O D=\frac{a \sqrt{3}}{6}=\frac{1}{3} \sqrt{S \sqrt{3} \cos \alpha}$.

From $\triangle O K D\left(\angle O K D=90^{\circ}\right): O K=O D \sin \angle O D K=\frac{\sin \alpha}{3} \sqrt{S \sqrt{3} \cos \alpha}$.

Answer: $\frac{\sin \alpha}{3} \sqrt{S \sqrt{3} \cos \alpha}$.