Let $M$ be the height of the cone, $R$ the radius of the base, $m$ the height of the cone protruding from the water, and $s$ the specific gravity to be found. Since the weight of the cone is equal to the weight of the displaced water, we have

$$
\frac{R^{2} \pi M}{3} s=\frac{R^{2} \pi M}{3}-\frac{R^{2} \pi}{3} \cdot \frac{m}{3}
$$

but

$$
M=R \sqrt{3}
$$

so

$$
\frac{R^{2} \pi R \sqrt{3}}{3} s=\frac{R^{2} \pi R \sqrt{3}}{3}-\frac{R^{2} \pi}{3} \cdot \frac{m}{3}
$$

Furthermore, since the areas of parallel sections are in the same ratio as the squares of the corresponding heights, we have

$$
3: 1=3 R^{2}: m^{2}
$$

from which

$$
m=R
$$

Substituting this into (1) and dividing by $R^{2} \pi$, we get

$$
\frac{\sqrt{3}}{3} s=\frac{\sqrt{3}}{3}-\frac{1}{9}
$$

from which

$$
s=1-\frac{\sqrt{3}}{3}
$$

or

$$
s=0.808
$$

(Ferencz Kräuter, Lugos.)

The problem was also solved by: E. Ádámffy, L. Bánó, I. Dömény, A. Földes, H. Friedländer, A. Fuchs, A. Haar, Z. Harsányi, Gy. Jánosy, G. Kertész, Gy. Krampera, I. Kürti, P. Messer, M. Pám, P. Rássy, M. Riesz, J. Rosenberg, Gy. Schwarz, J. Sonnenfeld, E. Tandlich, the V. district gymnasium mathematical circle.
We immerse an equilateral cone base-down in water. The circle where the water color touches the cone is one-third of the base circle's area, and the angle that the water color makes with the cone's side is $120^{\circ}$. What is the specific gravity of the material of the cone?