The density of the material of a compact, homogeneous sphere is $d$.

a) Cut out a small sphere from it such that when placed with its apex downward in water, the bounding conical surface is in contact with the water, while the spherical surface is in contact with the air. What is the opening angle of the cone? How deep is the apex of the small sphere from the water surface?

b) Also cut out a small sphere that floats such that its bounding spherical surface is in the water, and the conical surface is in contact with the air. Determine again the position of the apex of the small sphere relative to the water surface. (Stability of floating is not to be considered.)
a) Let the radius of the sphere be $r$ and the height of the spherical cap be $m$. Then the height of the cone is $r-m$, and the radius of its base is $\varrho^{2}=r^{2}-(r-m)^{2}=m(2 r-m)$. In the desired floating position, the mass of the water displaced by the cone and the mass of the spherical cap are equal:

$$
\begin{gathered}
\frac{m(2 r-m)(r-m) \pi}{3}=\frac{2 r \pi m \cdot r \cdot d}{3} \\
(2 r-m)(r-m)=2 r^{2} d
\end{gathered}
$$

From this, we can immediately get an equation for the desired depth $r-m=x$:

$$
(r+x) x=2 r^{2} d, \quad x^{2}+r x-2 r^{2} d=0
$$

The product of the two roots is: $-2 r^{2} d$, which is negative because $d>0$, so the two roots have opposite signs (and are certainly real). We can only use the positive root, so

$$
x=r(\sqrt{1+8 d}-1) / 2
$$

Floating is clearly only possible for $d<1$, and for $d>1$ it is below, with the distance from the surface always being $|r-m|$. The half-opening angle of the dense spherical cap can be calculated using the formula $\cos \alpha=(r-m) / r$. In the case of $d>1$, the floating is impossible, which some contestants did not consider, for example, the floating of iron ships.

However, some thought of a concave spherical cap in part a), where the entire body is above the outer water level, the conical cavity is filled with water, and thus the "contact" of the hull with the water "occurs". This is physically impossible.

Many contestants casually used the "arc cos" notation. Here "arc" = arc, meaning arc (paper arc), or the angle in radians. The "arc cos $x$" notation, however, means the "arc of the cosine," more precisely: the arc corresponding to the cosine value $x$. (This casualness was not considered a mistake.)

3. The equation (2) differs from the expanded form of (1) only in the term independent of $m$: $m^{2}-3 r m+2(1-d) r^{2}=0$. This can also be stated as follows: by substituting $(1-d)$ for $d$ in (1), we get (2), and from a physical standpoint, it can be interpreted as follows. When a body of volume $V$ and density $d$ is placed in water, a volume $V_{1}=V d$ of it submerges in water and a volume $V_{2}=V(1-d)$ remains in the air. The floating positions in a) and b) (for a convex spherical cap) exactly swap the submerged and air parts: $V_{1}^{\prime}=V_{2}$ and $V_{2}^{\prime}=V_{1}$; this explains the $d, 1-d$ swap.

Bodoky Andrea (Budapest, Teleki B. lg. IV. o. t.) Simonovits Miklós (Budapest, Radnóti M. gyak. g. IV. o. t.)

4. The previous remark explains the accidental observation that the height of the spherical cap satisfying the conditions in a) and b) is the same for $d=0.5$. In this case, $\alpha=51.8^{\circ}$, so the spherical cap with a total opening angle of $103.6^{\circ}$ can float in both positions.

Demendy Zoltán (Budapest, Hengersor úti g. IV. o. t.)

5. According to the text of the problem, we did not consider the stability of the floating in the above.