4. A sphere is inscribed in a regular triangular prism, touching all three sides and both bases of the prism. Find the ratio of the surface areas of the sphere and the prism.
Solution. Let $a$ be the side of the equilateral triangle which is the base of the prism, and let $R$ be the radius of the inscribed sphere. From the condition of the problem, it follows that the radius of the inscribed circle in the base of the prism is $R$, i.e., $R=\frac{a \sqrt{3}}{6}$. Since the sphere touches both bases, the height $h$ of the prism is $2R$, i.e., $R=\frac{h}{2}$. Let $P_{1}$ and $P_{2}$ be the surface areas of the sphere and the prism, respectively. Then $P_{1}=4 \pi R^{2}, \text{and}$

$$
P_{2}=a^{2} \frac{\sqrt{3}}{4}+a^{2} \frac{\sqrt{3}}{4}+3 a h=\frac{a^{2} \sqrt{3}}{2}+3 a h=2 \frac{(2 \sqrt{3} R)^{2} \sqrt{3}}{4}+3 \cdot 2 \sqrt{3} R \cdot 2 R=18 R^{2} \sqrt{3}
$$

Thus,

$$
\frac{P_{1}}{P_{2}}=\frac{4 \pi R^{2}}{18 R^{2} \sqrt{3}}=\frac{2 \pi}{9 \sqrt{3}}
$$

## 4th year