Create a section on a sphere whose area is equal to one-third of the great circle's area, and on the intersecting surface, place a cone whose apex falls into the center of the sphere; if the volume of the cone is $480 \mathrm{~m}^{3}$, what will be the radius of the sphere?
Let the radius of the sphere be $R$, the radius of the section be $r$; the area of the section is:

$$
r^{2} \pi=\frac{R^{2} \pi}{3}
$$

from which

$$
r^{2}=\frac{R^{2}}{3}
$$

The height of the cone is

$$
m=\sqrt{R^{2}-r^{2}}=R \sqrt{\frac{2}{3}}
$$

The volume of the cone is

$$
V=\frac{R^{3} \pi}{9} \sqrt{\frac{2}{3}}
$$

from which

$$
\begin{aligned}
R=\sqrt[3]{\frac{9 V \sqrt{\frac{3}{2}}}{\pi}} & =\sqrt[3]{\frac{9 \cdot 480 \sqrt{6}}{2 \pi}}=6 \sqrt[3]{\frac{10 \sqrt{6}}{\pi}} \\
R & =11.90 \text{~m}
\end{aligned}
$$

(Goldziher Károly and Kármán Tivadar.)

The problem was also solved by: Beck F., Dénes A., Devecis M., Friedmann B., Orlowszky F., Petrogalli G., Porde Gy., Schiffer H., Spitzer Ö., Szabó I.