12.293. The base of the oblique prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| C C_{1}\right)$ is an isosceles triangle, where $A B=A C=a$ and $\angle C A B=\alpha$. The vertex $B_{1}$ of the upper base is equidistant from all sides of the lower base, and the edge $B_{1} B$ forms an angle $\beta$ with the plane of the base. Find the volume of the prism.
## Solution.

Draw a perpendicular $B_{1} O$ from point $B_{1}$ to the plane $A B C$ and perpendiculars $B_{1} F, B_{1} K, B_{1} E$ to the sides $A B, B C, A C$ of triangle $A B C$ (Fig. 12.155), $\angle B{ }_{1} B O=\beta$.
$O F, O K, O E$ are the projections of $B_{1} F, B_{1} K, B_{1} E$ on the plane $A B C, O F \perp A B$, $O K \perp B C, O E \perp A C$ and, since $B_{1} F=B_{1} K=B_{1} E$, then $O F=O K=O E \Rightarrow O$ is the center of the circle inscribed in $\triangle A B C$, and lies on its height, median, and bisector $A K$.

\[
\begin{aligned}
& \angle A B K=\frac{\pi}{2}-\angle B A K=\frac{\pi}{2}-\frac{\alpha}{2}=\frac{\pi-\alpha}{2} \\
& \angle O B K=\frac{1}{2} \angle A B K=\frac{\pi-\alpha}{4}
\end{aligned}
\]

In $\triangle A K B\left(\angle A K B=\frac{\pi}{2}\right): B K=A B \sin \angle B A K=a \sin \frac{\alpha}{2}$.

In $\triangle B K O\left(\angle B K O=\frac{\pi}{2}\right): B O=\frac{B K}{\cos \angle O B K}=\frac{a \sin \frac{\alpha}{2}}{\cos \frac{\pi-\alpha}{4}}$.

In $\triangle B_{1} O B\left(\angle B_{1} O B=\frac{\pi}{2}\right): B_{1} O=B O \operatorname{tg} \angle B_{1} B O=\frac{a \sin \frac{\alpha}{2} \operatorname{tg} \beta}{\cos \frac{\pi-\alpha}{4}}$.

The area of the base of the prism $S=\frac{1}{2} a^{2} \sin \alpha$, and its volume $V=S \cdot B_{1} O=$ $=\frac{a^{3} \sin \alpha \sin \frac{\alpha}{2} \operatorname{tg} \beta}{2 \cos \frac{\pi-\alpha}{4}}$.

Answer: $\frac{a^{3} \sin \alpha \sin \frac{\alpha}{2} \operatorname{tg} \beta}{2 \cos \frac{\pi-\alpha}{4}}$.