A sphere with radius $R$ is cut by two parallel planes such that the surface area of the spherical segment between them is constant, but the volume of the spherical layer is maximized!
I. Solution. The surface area of the spherical cap: $F=2 R \pi m$, where $m$ is the height of the spherical cap or the corresponding spherical layer. Since $F$ and $R$ are constant, $m$ must also be constant.

A spherical layer of constant height has the largest volume when the areas of the parallel plane sections (bounded by the planes) are the largest; this occurs when the plane of the largest circle of the sphere bisects the height of the spherical layer perpendicularly.

II. Solution. The radii of the circles bounding the spherical layer are $r_{1}$ and $r_{2}$. The volume is:

$$
V=\frac{1}{6} \pi m^{3}+\frac{1}{2} \pi m\left(r_{1}^{2}+r_{2}^{2}\right)
$$

Let $r$ be the radius of the cross-section at the perpendicular bisector of the distance $(m)$ between the bounding circles; then

$$
r_{1}^{2}+r_{2}^{2}=2 r^{2}-\frac{1}{2} m^{2}
$$

The value of $V$ is largest when $r_{1}^{2}+r_{2}^{2}$ is the largest; this occurs when $r=R$.

$$
V_{\max }=\frac{1}{6} \pi m^{3}+\frac{1}{2} \pi m\left(2 R^{2}-\frac{1}{2} m^{2}\right)=\pi m R^{2}-\frac{1}{12} \pi m^{3}
$$

Note. If $m=2 R$, then this volume value becomes $\frac{4}{3} \pi R^{3}$!

III. Solution. 

![](https://cdn.mathpix.com/cropped/2024_05_02_501d1d79198427a0ad3cg-1.jpg?height=410&width=415&top_left_y=1020&top_left_x=844)

In the previous solution, we saw that the volume of the spherical layer depends on the sum $r_{1}^{2}+r_{2}^{2}$. Consider the distance $x$ from the endpoint of the diameter perpendicular to the circle with radius $r_{2}$ as an independent variable. The distance from the same endpoint to the circle with radius $r_{1}$ is $x+m$. Therefore, considering that $r_{1}$ and $r_{2}$ are the geometric means of the corresponding segments of the diameter,

$$
\begin{gathered}
r_{2}^{2}=x(2 R-x), \quad r_{1}^{2}=(x+m)[2 R-(x+m)] \\
r_{1}^{2}+r_{2}^{2}=-2 x^{2}+(4 R-2 m) x+2 R m-m^{2}
\end{gathered}
$$

Thus, $r_{1}^{2}+r_{2}^{2}$ is a quadratic function of $x$, which has a maximum when

$$
x=\frac{4 R-2 m}{4}=R-\frac{m}{2}
$$

Therefore,

$$
x+m=R+\frac{m}{2}
$$

The distances from the center of the sphere to the planes bounding the spherical layer are equal to $\frac{m}{2}$.

Hamza Aladár (Kegyesrendi g. VIII. o. Szeged.)