A cone with a weight of $s$, whose height is $m$, and the radius of its base is $r$, is immersed in a liquid with a weight of $s^{\prime}$, base down. To what height will the cone sink in the liquid?
If the height of the submerged cone is $x$ and the radius of its base is $r_{1}$, then

$$
x: r_{1}=m: r
$$

from which

$$
r_{1}=\frac{r x}{m}
$$

The volume of the whole cone is

$$
K=\frac{r^{2} \pi m}{3}
$$

and that of the submerged part is

$$
K_{1}=\frac{r^{2} \pi m}{3}-\frac{r^{2} x^{3} \pi}{3 m^{2}}
$$

According to the condition,

$$
\frac{r^{2} \pi m s}{3}=s^{\prime}\left(\frac{r^{2} \pi m}{3}-\frac{r^{2} x^{3} \pi}{3 m^{2}}\right)
$$

or

$$
m s=m s^{\prime}-\frac{x^{3} s^{\prime}}{m^{2}}
$$

from which

$$
x=m \sqrt[3]{\frac{s^{\prime}-s}{s^{\prime}}}
$$

and thus the height of the submerged frustum of the cone is

$$
(m-x)=m\left(1-\sqrt[3]{\frac{s^{\prime}-s}{s^{\prime}}}\right)
$$

(Krisztián György.)

The problem was also solved by: Barabás S., Csete F., Dolowschiák M., Filkorn J., Freibauer E., Juvancz I., Kohn B., Kornis Ö., Krausz B., Lukhaub Gy., Obláth R., Oltay K., Pollák N., Rehberger Z., Sasvári G., Spitzer Ö., Weisz J.