A hemisphere with radius $r=1$ is to be divided into two equal volume parts by a plane parallel to the base. How high is each of the two parts?
Let the height of the spherical cap be $x$. According to the problem, the volume of the spherical cap is half the volume of a hemisphere, that is,

$$
\frac{\pi x^{2}}{3}(3 r-x)=\frac{\pi}{3} r^{3}
$$

If we rearrange the equation and consider that $r=1$, we get

$$
x^{3}-3 x^{2}+1=0
$$

Let $x=y+1$, then from (2) we obtain the following equation:

$$
y^{3}-3 y-1=0
$$

Since in this case

$$
\left(-\frac{1}{2}\right)^{2}+\left(-\frac{3}{3}\right)^{3}=-\frac{3}{4}<0
$$

the roots of the equation are real and have the following values:

$$
\begin{gathered}
y_{1}=2 \cos \frac{\alpha}{3}=2 \cos 20^{\circ} \\
y_{2}=2 \cos 140^{\circ}=-2 \cos 40^{\circ} \\
y_{3}=2 \cos 260^{\circ}=-2 \cos 80^{\circ}
\end{gathered}
$$

However, it is necessary that $1<x<1$, and therefore the only value that satisfies the problem is $y_{3}=2 \cos 80^{\circ}=-0.3473$, which gives:

$$
x=y+1=0.6527
$$

Thus, the height of the spherical cap is 0.6527, and the height of the spherical segment is 0.3473.

(Bernát Friedmann.)

The problem was also solved by: Mihály Devecis and István Szabó.

```
\({ }^{1}\) It is known from the theory of cubic equations that if
                \(x^{3}+a x+b=0\)
and
                \(\left(\frac{b}{2}\right)^{2}+\left(\frac{a}{3}\right)^{3} \leqq 0\)
then the roots of the equation are:

\[
\begin{gathered}
x_{1}=2 \sqrt{-\frac{a}{3}} \cos \alpha, x_{2}=2 \sqrt{-\frac{a}{3}} \cos \left(\frac{\alpha}{3}+120^{\circ}\right) \\
x_{3}=2 \sqrt{-\frac{a}{3}} \cos \left(\frac{\alpha}{3}+240^{\circ}\right)
\end{gathered}
\]
where

\[
\cos \alpha=\frac{-\frac{b}{2}}{\left(\sqrt{-\frac{a}{3}}\right)^{3}}
\]```