12.247. The angle between the height and the slant height of a cone is $\alpha$. A regular triangular prism is inscribed in the cone; the lower base of the prism lies in the plane of the base of the cone. The lateral faces of the prism are squares. Find the ratio of the lateral surfaces of the prism and the cone.
## Solution.

Let $S$ be the vertex, $O$ the center of the base of the given cone, $S A$ the generatrix of the cone passing through the vertex $C$ of the inscribed regular prism (Fig. 12.113), $\angle A S O=\alpha$. The lateral edge $C B$ of the prism is parallel to $S O$, so $\angle A C B=\angle A S O=\alpha$.

If the edge of the prism is 1, then the area of its lateral surface $S_{1}=3, O B=\frac{\sqrt{3}}{3} \cdot \mathrm{B} \triangle C B A\left(\angle C B A=\frac{\pi}{2}\right): B A=C B \operatorname{tg} \angle A C B=\operatorname{tg} \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0996.jpg?height=854&width=537&top_left_y=78&top_left_x=466)

Fig. 12.113

$O A=O B+B A=\frac{\sqrt{3}}{3}+\operatorname{tg} \alpha=\operatorname{tg} \frac{\pi}{6}+\operatorname{tg} \alpha=\frac{\sin \left(\alpha+\frac{\pi}{6}\right)}{\cos \frac{\pi}{6} \cos \alpha}=\frac{2 \sin \left(\alpha+\frac{\pi}{6}\right)}{\sqrt{3} \cos \alpha}$.

In $\triangle S O A\left(\angle S O A=\frac{\pi}{2}\right): S A=\frac{O A}{\sin \angle A S O}=\frac{2 \sin \left(\alpha+\frac{\pi}{6}\right)}{\sqrt{3} \cos \alpha \sin \alpha}=\frac{4 \sin \left(\alpha+\frac{\pi}{6}\right)}{\sqrt{3} \sin 2 \alpha}$.

The lateral surface of the cone $S_{2}=\pi \cdot O A \cdot S A=\frac{8 \pi \sin ^{2}\left(\alpha+\frac{\pi}{6}\right)}{3 \sin 2 \alpha \cos \alpha}$.

Then $\frac{S_{1}}{S_{2}}=\frac{9 \sin 2 \alpha \cos \alpha}{8 \pi \sin ^{2}\left(\alpha+\frac{\pi}{6}\right)}$.

Answer: $\frac{9 \sin 2 \alpha \cos \alpha}{8 \pi \sin ^{2}\left(\alpha+\frac{\pi}{6}\right)}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0997.jpg?height=597&width=724&top_left_y=63&top_left_x=284)

Fig. 12.114