The lateral surface area of a certain cone is $p$ square units. The angle at the vertex is $\alpha$. What is the volume? ( $p=160 \mathrm{dm}^{2}$, $\alpha=60^{\circ} 20^{\prime}$.)
We denote the slant height of the given cone by $l$, the radius of its base circle by $r$, and its height by $m$. Then,

$$
r=\operatorname{lsin} \frac{\alpha}{2}
$$

and

$$
l r \pi=p
$$

thus,

$$
r=\sqrt{\frac{p \sin \frac{\alpha}{2}}{\pi}}
$$

Furthermore,

$$
m=r \operatorname{ctg} \frac{\alpha}{2}=\cos \frac{\alpha}{2} \sqrt{\frac{p}{\pi \sin \frac{\alpha}{2}}}
$$

and hence the volume in question is

$$
K=\frac{r^{2} m \pi}{3}=\frac{r m \pi}{3} \cdot r=\frac{p \cos \frac{\alpha}{2}}{3} \sqrt{\frac{p \sin \frac{\alpha}{2}}{\pi}}
$$

Substituting the given numerical values, we obtain

$$
K=233.27 \mathrm{dm}^{3}
$$

(István Losonczy, Fiume.)

Number of solutions: 26.