In a sphere of radius $R$, we inscribe an equilateral cone. We place a plane parallel to the base of the cone. At which position of the plane will the sum of the cross-sections of the cone and the sphere be maximal?
Let the distance of the cutting plane from the center of the sphere be $x$; the radius of the circle cut from the sphere be $r_{1}$, and that cut from the cone be $r_{2}$. Since the radius of the base of the equilateral cone inscribed in the sphere is $\frac{R}{2} \sqrt{3}$, and its height is $\frac{3}{2} R$, we have:

$$
\frac{3}{2} R:(R+x)=\frac{R}{2} \sqrt{3}: r_{2}
$$

from which

$$
r_{2}^{2}=\frac{(R+x)^{2}}{3} ; \text { but } r_{1}^{2}=R^{2}-x^{2}
$$

and thus the sum of the areas of the sections is:

$$
t=\left(r_{1}^{2}+r_{2}^{2}\right) \pi=\left[R^{2}-x^{2}+\frac{(R+x)^{2}}{3}\right]=\frac{2 \pi}{3}\left[-x^{2}+R x+2 R^{2}\right]
$$

This function attains its maximum value when $x=\frac{R}{2}$, that is, when the cutting plane coincides with the base of the cone.

(Béla Bayer, Losoncz.)

Number of solutions: 32.