A right circular cone is floating in water, its specific gravity is $s$, the angle at the vertex of its axial section is $\alpha$; the height of the part above water is $m$; what is the height of the part submerged in water and what is the weight of the whole cone?
I. solution. The weight of the cone is equal to the weight of the displaced water, thus:

$$
\frac{1}{3} \pi s(m+x)^{3} \operatorname{tg}^{2} \frac{1}{2} \alpha=\frac{1}{3} \pi x\left[(m+x)^{2} \operatorname{tg}^{2} \frac{1}{2} \alpha+m^{2} \operatorname{tg}^{2} \frac{1}{2} \alpha++m(m+x) \operatorname{tg}^{2} \frac{1}{2} \alpha\right]
$$

from which

$$
\begin{gathered}
(m+x)^{3} \cdot s=(m+x)^{3}-m^{3} \\
(m+x)^{3}=\frac{m^{3}}{1-s} \\
x=m\left[\sqrt[3]{\frac{1}{1-s}}-1\right]
\end{gathered}
$$

The weight of the cone is

$$
Q=\frac{(m+x)^{3} \pi s \operatorname{tg}^{2} \frac{1}{2} \alpha}{3}
$$

Substituting (2) into (4), we get:

$$
Q=\frac{\pi m^{3} s}{3(1-s)} \operatorname{tg}^{2} \frac{1}{2} \alpha
$$

(János Galter, grammar school VIII. grade, Sz-Udvarhely.)

II. solution. Let $V$ be the volume of the entire cone, and $V_{1}$ be the volume of the submerged part, then:

$$
V: V_{1}=1: s
$$

from which

$$
V:\left(V-V_{1}\right)=1:(1-s)
$$

But $V-V_{1}$ is the volume of the part of the cone that protrudes above the water; the volume of this part is to the volume of the entire cone as the cubes of the corresponding heights; thus

$$
M^{3}: m^{3}=1:(1-s)
$$

from which

$$
M=\frac{m}{\sqrt[3]{1-s}}
$$

and from this

$$
x=M-m=m\left[\sqrt[3]{\frac{1}{1-s}}-1\right]
$$

(Bernát Friedmann, grammar school VII. grade, S.-A.-Ujhely).

The problem was also solved by: Gusztáv Szabó, Győr; Ervin Hofbauer and Gyula Pósch, Budapest; Leó Goldberger, Béla Grünhut and Aladár Visnya, Pécs.