We erect a straight cone over the base of a hemisphere with radius $r$, the height of which is 2r. Calculate the volume of the common solid formed by the two bodies.
Let $D F=a, F G=m$.

The volume $V$ of the common part of the two bodies is equal to the sum of the volumes of a frustum of a cone and a spherical cap.

![](https://cdn.mathpix.com/cropped/2024_05_02_015eb460adc816825e04g-1.jpg?height=358&width=415&top_left_y=255&top_left_x=844)

The volume of the frustum of the cone:

$$
V_{1}=\frac{2}{3} r^{3} \pi-\frac{2}{3} a^{3} \pi=\frac{2}{3} \pi\left(r^{3}-a^{3}\right)
$$

From triangle DFO

$$
r^{2}=a^{2}+(2 r-2 a)^{2}
$$

or

$$
5 a^{2}-8 a r+3 r^{2}=0
$$

from which

$$
a=\frac{4 \pm 1}{5} r
$$

Out of the two values of $a$, only one can be used because, according to the problem, $a$ cannot be equal to $r$; thus, $a=\frac{3}{5} r$. Substituting this value of $a$ into (1),

$$
V_{1}=\frac{2}{3} \pi\left(r^{3}-\frac{27}{125} r^{3}\right)=\frac{196}{375} r^{3} \pi
$$

The volume of the spherical cap:

$$
V_{2}=m^{2} \pi\left(r-\frac{m}{3}\right)
$$

where

and thus

$$
m=2 a-r=\frac{6}{5} r-r=\frac{r}{5}
$$

$$
V_{2}=\frac{r^{2} \pi}{25}\left(r-\frac{r}{15}\right)=\frac{14 r^{3} \pi}{375}
$$

Adding the volumes of the frustum of the cone and the spherical cap, we get the desired volume;

$$
\begin{aligned}
& V=\frac{196}{375} r^{3} \pi+\frac{14}{375} r^{3} \pi \\
& V=\frac{210}{375} r^{3} \pi=\frac{14}{25} r^{3} \pi
\end{aligned}
$$

(Jenö Fekete.)

The problem was also solved by: Bojedain F., Devecis M., Fleischmann S., Petrogalli G., Probst E., Spitzer Ö., Szabó I., Szabó K.