A cone is inscribed in a sphere; at what distance from the base of the cone should a plane parallel to the base be placed so that the area of the circle cut from the sphere is twice the area of the circle cut from the cone? The radius of the base of the cone is $r$, and the height of the cone is $m$.
Let the sought distance be $D G=x$, the radius of the circle cut from the sphere be $E G=y$, and the radius of the circle cut from the cone be $F G=z$.

![](https://cdn.mathpix.com/cropped/2024_05_02_cf26cd8ae57b145b8127g-1.jpg?height=401&width=398&top_left_y=190&top_left_x=842)

According to the problem:

$$
y^{2} \pi=2 z^{2} \pi
$$

or

$$
y^{2}=2 z^{2}
$$

From the similar triangles $A F G$ and $A B D$:

$$
z=\frac{r(m-x)}{m}
$$

Substituting this into (1):

$$
y^{2}=\frac{2 r^{2}(m-x)^{2}}{m^{2}}
$$

If we denote the radius of the sphere by $R$, then from the triangle $B O D$:

$$
R^{2}=r^{2}+(m-R)^{2}
$$

from which

$$
R=\frac{r^{2}+m^{2}}{2 m}
$$

From the triangle $EGO$:

$$
\begin{aligned}
& y^{2}=R^{2}-[R-(m-x)]^{2} \\
& =2(m-x) R-(m-x)^{2}
\end{aligned}
$$

Substituting (3) and (4) into this and canceling $(m-x)$:

$$
\frac{2 r^{2}(m-x)}{m^{2}}=2 \frac{r^{2}+m^{2}}{2 m}-(m-x)
$$

from which

$$
m-x=\frac{m r^{2}+m^{3}}{2 r^{2}+m^{2}}
$$

and thus

$$
x=\frac{m r^{2}}{2 r^{2}+m^{2}}
$$

(Jenő Fekete.)

The problem was also solved by: Bojedain F., Détshy K., Devecis M., Kornis Ö., Lukhaub Gy., Magyar S., Petrogalli G., Probst E., Spitzer Ö., Szabó I.