1. The height of a right cone is twice the length of the radius of its base. Find the ratio of the volume of the sphere circumscribed around the cone to the sphere inscribed in it.
Solution. Let $R=\overline{O V}$ be the radius and $O$ be the center of the circumscribed sphere around the cone, point $N$ is the midpoint of the base of the cone, point $V$ is the vertex of the cone, $r=\overline{S N}$ is the radius and $S$ is the center of the inscribed sphere in the cone, $v=\overline{V N}$ is the height of the cone, $\rho$ is the radius of the base, $s$ is the length of the lateral side.

Then,

$$
\begin{aligned}
& s=\sqrt{4 \rho^{2}+\rho^{2}}=\rho \sqrt{5}, P_{\triangle A B V}=2 \rho^{2} \\
& R=\frac{\overline{A B} \cdot \overline{B V} \cdot \overline{V A}}{4 P_{\triangle A B V}}=\frac{2 \rho \cdot \rho \sqrt{5} \cdot \rho \sqrt{5}}{8 \rho^{2}}=\frac{5 \rho}{4} \\
& r=\frac{2 P_{\triangle A B V}}{L}=\frac{4 \rho^{2}}{2 \rho+2 \sqrt{5} \rho}=\frac{2 \rho}{1+\sqrt{5}}
\end{aligned}
$$

Therefore, the desired ratio is

$$
\begin{aligned}
\frac{V_{R}}{V_{r}} & =\frac{\frac{4}{3} R^{3} \pi}{\frac{4}{3} r^{3} \pi}=\left(\frac{R}{r}\right)^{3}=\left(\frac{\frac{5 \rho}{4}}{\frac{2 \rho}{1+\sqrt{5}}}\right)^{3} \\
& =\left(\frac{5(1+\sqrt{5})}{8}\right)^{3}=\frac{125(2+\sqrt{5})}{64}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_39af436b8d125b2f3e1dg-5.jpg?height=540&width=479&top_left_y=475&top_left_x=991)