A compact sphere of specific gravity $s$ floats in water such that only the spherical cap of the sphere protrudes from the water. What is the half-angle of the cone of the spherical cap? What is the condition for solvability? What is the half-angle if $s=0.75?$ (Half-angle: the angle between the generator and the axis of rotation.)





Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
I. Solution: a) According to the well-known law of Archimedes, the weight of the spherical segment equals the weight of the water displaced by the submerged volume (in this case, the frustum of a cone). If we denote the volume of the spherical segment by $K$ and the volume of the frustum of a cone by $K^{\prime}$, then (since the specific weight of water is 1)

$$
K \cdot s=K^{\prime} \cdot 1
$$

Our diagram shows the axial section of the spherical segment and the labeling.

![](https://cdn.mathpix.com/cropped/2024_05_02_f908a888ce3dbf315235g-1.jpg?height=501&width=504&top_left_y=449&top_left_x=797)

Accordingly,

$$
K=\frac{2 r^{2} \pi(r-m)}{3}, \quad K^{\prime}=\frac{\varrho^{2} \pi m}{3}
$$

Substituting these values into (1) and simplifying by $\frac{\pi}{3}$,

$$
2 r^{2}(r-m) s=\varrho^{2} m
$$

If we express the values of $\varrho$ and $m$ in terms of $r$ and the half-angle $x$ of the frustum of a cone:

$$
\varrho=r \sin x, \quad m=r \cos x
$$

and thus $r-m=r(1-\cos x)$, we get

$$
2 s r^{3}(1-\cos x)=r^{3} \sin ^{2} x \cos x
$$

Substituting $\sin ^{2} x$ with $1-\cos ^{2} x=(1-\cos x)(1+\cos x)$ and simplifying both sides by $r^{3}(1-\cos x)$ (since $0<x<\frac{\pi}{2}$, $1-\cos x \neq 0$),

$$
2 s=\cos x+\cos ^{2} x, \quad \text { or } \quad \cos ^{2} x+\cos x-2 s=0
$$

Since $0<x<\frac{\pi}{2}$, only the positive root makes sense. Therefore,

$$
\cos x=\frac{-1+\sqrt{1+8 s}}{2}
$$

b) The problem is solvable if

$$
0<\frac{\sqrt{1+8 s}-1}{2}<1
$$

which implies

$$
1<\sqrt{1+8 s}<3, \quad \text { or } \quad 0<s<1
$$

Thus, the solvability condition coincides with the general floating condition known from physics.
c) For $s=0.75$,

$$
\cos x=\frac{\sqrt{7}-1}{2} \approx 0.8229, \quad \text { and thus } \quad x=34^{\circ} 37^{\prime}
$$

Argay Gyula (Balassagyarmat, Balassi B. g. III. o. t.)

II. Solution: Based on the Pythagorean theorem,

$$
\varrho^{2}=r^{2}-m^{2}
$$

Substituting this value into (2) and simplifying by $(r-m)$,

$$
2 r^{2} s=(r+m) m
$$

Rearranging for $m$, we get the following quadratic equation

$$
m^{2}+r m-2 r^{2} s=0
$$

The positive root of this equation is

$$
m=\frac{\sqrt{1+8 s}-1}{2} r
$$

The solvability condition in this case is

$$
0<\frac{\sqrt{1+8 s}-1}{2} r<r
$$

which again yields the condition $0<s<1$.

For the half-angle of the cone,

$$
\cos x=\frac{m}{r}=\frac{\sqrt{1+2 s}-1}{2}
$$

is obtained, as in the first solution.

Pogány Eörs (Bp., V., Eötvös J. g. III. o. t.)