The volume of a frustum circumscribed around a sphere is twice the volume of the inscribed sphere. How many times larger is the radius of the base circle of the frustum compared to the radius of the top circle?
We know that the volume of a frustum of a cone is

$$
V_{\mathrm{cs}}=\frac{\pi}{3} m\left(R^{2}+r^{2}+R r\right)
$$

where $R$ is the radius of the base circle, and $r$ is the radius of the top circle. The volume of a sphere is:

$$
V_{\mathrm{g}}=\frac{4 \pi \varrho^{3}}{3}
$$

where $\varrho$ is the radius of the sphere. According to the condition,

$$
\frac{\pi}{3} m\left(R^{2}+r^{2}+R r\right)=2 \cdot \frac{4 \pi \varrho^{3}}{3}
$$

Simplifying and substituting $m=2 \varrho$, we get

$$
R^{2}+r^{2}+R r=4 \varrho^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_659e03da92392c63d6f3g-1.jpg?height=604&width=526&top_left_y=774&top_left_x=794)

We know that if the frustum of the cone is cut by a plane passing through the center of the base circle and perpendicular to the base plane, the cross-section is an isosceles trapezoid. The parallel sides of the trapezoid are $2 R$ and $2 r$, and the legs are $R+r$ due to the equality of the tangent segments.

According to the Pythagorean theorem,

$$
(2 \varrho)^{2}=(R+r)^{2}-(R-r)^{2}=4 R r
$$

Substituting this into (1), we get

$$
R^{2}+r^{2}-3 R r=0
$$

Dividing the equation by $R r \neq 0$ and introducing the new variable $\frac{R}{r}=u$, we get the following quadratic equation after simplification:

$$
u^{2}-3 u+1=0
$$

from which

$$
u=\frac{R}{r}=\frac{3+\sqrt{5}}{2}, \quad \text { i.e., } \quad R=\frac{3+\sqrt{5}}{2} r \approx 2.618
$$

(The other root of the equation gives the value of $\frac{r}{R}$.)