A given radius sphere is to be enclosed by a frustum of a cone, whose volume is twice that of the sphere. The radii of the base and top of the frustum, as well as the radius of the circle along which the mantle of the frustum touches the sphere, are to be determined.
Let's denote the radius of the sphere by $R$, the radii of the bases of the truncated cone by $r$ and $r'$, and the radius of the circle along which the sphere and the truncated cone touch by $\zeta$.

According to the problem,

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

or

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

Furthermore, the slant height of the truncated cone is $l=r+r^{\prime}$, and thus

$$
\left(r+r^{\prime}\right)^{2}=4 R^{2}+\left(r-r^{\prime}\right)^{2}
$$

or

$$
R^{2}=r r^{\prime}
$$

If we add equation 1) to equation 2), we get the following equation:

$$
r+r^{\prime}=R \sqrt{5}
$$

and if we subtract three times equation 2) from equation 1), we get:

$$
r-r^{\prime}=R
$$

From equations 3) and 4), it follows that:

$$
\begin{aligned}
r & =\frac{R}{2}(\sqrt{5}+1) \\
r^{\prime} & =\frac{R}{2}(\sqrt{5}-1)
\end{aligned}
$$

Finally,

$$
\left(r-r^{\prime}\right):\left(r+r^{\prime}\right)=\left(\zeta-r^{\prime}\right): r^{\prime}
$$

and from this,

$$
\begin{gathered}
\zeta=\frac{2 r r^{\prime}}{r+r^{\prime}}=\frac{2 R^{2}}{R \sqrt{5}} \\
\zeta=\frac{2 R \sqrt{5}}{5}
\end{gathered}
$$

Kutlik Igor, Pozsony; Seidner Mihály, Losoncz.