12.319. Find the ratio of the volume of a regular $n$-sided pyramid to the volume of the circumscribed sphere, if the angle between the lateral edge and the plane of the base of the pyramid is $\alpha$.
Solution.

Let $\mathrm{CO}_{2}$ be the height of a regular $n$-sided pyramid, and $A_{1} A_{2}$ be the side of its base. Then $\angle C A_{1} O_{2}=\alpha, \angle A O_{1} B=\frac{2 \pi}{\mathrm{n}}$.

The center $\mathrm{O}_{1}$ of the circumscribed sphere is the intersection of the segment $\mathrm{CO}_{1}$ (Fig. 12.180) or its extension beyond point $O_{1}$ (Fig. 12.180, a) and the perpendicular bisector $O_{1} B$ of the lateral edge $A_{1} C$ of the pyramid.
$\triangle A_{1} O_{2} C$ and $\triangle O, B C$ are right triangles with a common acute angle at vertex $C$. Then $\angle B O_{1} C=\angle C A_{1} O_{2}=\alpha$.

If the radius of the sphere $O_{1} C=R$, then from $\triangle C B O_{1}\left(\angle C B O_{1}=\frac{\pi}{2}\right): C B=$ $=C O_{1} \sin \angle B O_{1} C=R \sin \alpha$.

$A_{1} C=2 C B=2 R \sin \alpha$.

In $\triangle A_{1} O_{2} C\left(\angle A_{1} O_{2} C=\frac{\pi}{2}\right): C O_{2}=A C \sin \angle C A_{1} O_{2}=2 R \sin ^{2} \alpha$;

$A_{1} O_{2}=A_{1} C \cos \angle C A_{1} O_{2}=2 R \sin \alpha \cos \alpha=R \sin 2 \alpha$.

The area of the base of the pyramid:

$$
S=n S_{\triangle A_{1} O_{2} A_{2}}=\frac{n}{2} A_{1} O_{2}^{2} \sin \angle A_{1} O_{2} A_{2}=\frac{n}{2} R^{2} \sin ^{2} 2 \alpha \sin \frac{2 \pi}{n}
$$

The volume of the pyramid $V_{1}=\frac{1}{3} S \cdot \mathrm{CO}_{2}=\frac{1}{3} n R^{3} \sin ^{2} \alpha \sin ^{2} 2 \alpha \sin \frac{2 \pi}{n}$, the volume of the sphere $V_{2}=\frac{4}{3} \pi R^{3}$ and $\frac{V_{1}}{V_{2}}=\frac{n \sin ^{2} \alpha \sin ^{2} 2 \alpha \sin \frac{2 \pi}{n}}{4 \pi}$.

$$
\text { Answer: } \frac{n \sin ^{2} \alpha \sin ^{2} 2 \alpha \sin \frac{2 \pi}{n}}{4 \pi}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1075.jpg?height=624&width=571&top_left_y=1214&top_left_x=94)

Fig. 12.181