12.295. The base of a right prism described around a sphere of radius $r$ is a right triangle with an acute angle $\alpha$. Find the volume of the prism.
## Solution.

Let $\triangle A B C$ be the lower base of the given right prism, $\angle A C B=90^{\circ}, \angle B A C=\alpha$ (Fig. 12.157), the height of the prism $H=2 r$, the projection of the sphere onto the plane $A B C$ is a circle inscribed in $\triangle A B C$, with radius $r$, $O$ is the center of this circle, $E$ and $F$ are the points of tangency of the circle with the legs $A C$ and $B C$.

Then $O E=O F=E C=C F=r, \angle O A E=\frac{1}{2} \angle B A C=\frac{\alpha}{2}, O E \perp A C$.

In $\triangle A E O\left(\angle A E O=\frac{\pi}{2}\right): A E=r \operatorname{ctg} \frac{\alpha}{2}$.

$A C=A E+E C=r\left(1+\operatorname{ctg} \frac{\alpha}{2}\right)$.

In $\triangle A C B\left(\angle A C B=\frac{\pi}{2}\right): B C=A C \operatorname{tg} \alpha=r\left(1+\operatorname{ctg} \frac{\alpha}{2}\right) \operatorname{tg} \alpha$.

The area of the base of the prism

$$
S=\frac{1}{2} A C \cdot B C=\frac{1}{2} r^{2}\left(1+\operatorname{ctg} \frac{\alpha}{2}\right)^{2} \operatorname{tg} \alpha=\frac{1}{2} r^{2}\left(\frac{\sin \frac{\alpha}{2}+\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}\right)^{2} \operatorname{tg} \alpha=
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1050.jpg?height=577&width=771&top_left_y=77&top_left_x=299)

Fig. 12.158

$$
\begin{aligned}
& =\frac{1}{2} r^{2} \cdot \frac{1+\sin \alpha}{\sin ^{2} \frac{\alpha}{2}} \cdot \frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\cos \alpha}=r^{2} \cdot \frac{1+\sin \alpha}{\cos \alpha} \operatorname{ctg} \frac{\alpha}{2}= \\
& =r^{2} \cdot \frac{1+\cos \left(\frac{\pi}{2}-\alpha\right)}{\sin \left(\frac{\pi}{2}-a\right)} \operatorname{ctg} \frac{\alpha}{2}=r^{2} \operatorname{ctg} \frac{\alpha}{2} \operatorname{ctg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)
\end{aligned}
$$

The volume of the prism $V=S \cdot H=2 r^{3} \operatorname{ctg} \frac{\alpha}{2} \operatorname{ctg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$

Answer: $2 r^{3} \operatorname{ctg} \frac{\alpha}{2} \operatorname{ctg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$