An empty cone of height $m$ stands on its vertex, with its axis in a vertical position. The cone is filled with water to a height of $a$ dm from the vertex; if we add $p \mathrm{~d} m^{3}$ of water, the water level rises by $n$ dm. What is the volume of the cone and the radius of its base?
The added water fills a frustum of a cone, whose volume, if the angle at the vertex is $\alpha$

$$
p=\frac{\pi(a+n)^{3} \operatorname{tg}^{2} \frac{\alpha}{2}}{3}-\frac{\pi a^{3} \operatorname{tg}^{2} \frac{\alpha}{2}}{3}
$$

from this

$$
\operatorname{tg} \frac{\alpha}{2}=\sqrt{\frac{3 p}{\pi\left[(a+n)^{3}-a^{3}\right]}}
$$

therefore, the radius of the base of the cone:

$$
r=m \operatorname{tg} \frac{\alpha}{2}=m \sqrt{\frac{3 p}{\pi\left[(a+n)^{3}-a^{3}\right]}}
$$

and

$$
k=\frac{r^{2} \pi m}{3}=\frac{m^{3} p}{(a+n)^{3}-a^{3}}
$$

(Imre Bartók, Budapest.)

Number of solutions: 36.