A sphere of radius $R$ and a circular cone, whose base circle has a radius of $R$ and height $2R$, rest on the same plane. We cut these two bodies with a plane parallel to the base plane. How far from the base plane must the cutting plane be so that

1) the areas of the circles cut from the two bodies are equal to each other;
2) the volume of the frustum of the cone between the base and the cutting plane is $n$ times the volume of the spherical cap between the same planes.

For which values of $n$ is the problem possible?
$1^{\circ}$. Let the distance of the intersecting plane from the base plane be $m$, and the common value of the radii of the intersected circles be $r$. Determining the value of $r$ from the sphere and the cone, we get the following equations:

$$
\begin{gathered}
r^{2}=R^{2}-(R-m)^{2}, \quad r=R \frac{2 R-m}{2 R} \\
r^{2}=2 R m-m^{2}, \quad 2 r=2 R-m
\end{gathered}
$$

or by eliminating $r$ from the two equations,

$$
\begin{gathered}
(2 R-m)^{2}=8 R m-4 m^{2}=4 R^{2}-4 R m+m^{2} \\
5 m^{2}-12 R m+4 R^{2}=0 \\
10 m=12 R \pm \sqrt{144 R^{2}-80 R^{2}} \\
10 m=12 R \pm 8 R \\
m_{1}=\frac{2}{5} R \\
m_{2}=2 R
\end{gathered}
$$

$2^{\circ}$. Keeping the above notations, the volume of the frustum of the cone is

$$
V=\frac{m \pi}{3}\left(R^{2}+R r+r^{2}\right)
$$

the volume of the spherical cap is

$$
V^{\prime}=\frac{m^{2} \pi}{3}(3 R-m)
$$

while again

$$
2 r=2 R-m
$$

so the new form of $V$ is

$$
V=\frac{m \pi}{12}\left(m^{2}+6 R m+12 R^{2}\right)
$$

and thus, according to the condition,

$$
\begin{gathered}
\frac{m \pi}{12}\left(m^{2}-6 R m+12 R^{2}\right)=n \frac{m^{2} \pi}{3}(3 R-m) \\
m^{2}-6 R m+12 R^{2}=4 n m(3 R-m) \\
(4 n+1) m^{2}-6 R(2 n+1) m+12 R^{2}=0
\end{gathered}
$$

For the problem to be solvable, it is necessary that the roots of 1) are real and at least one of them is less than $2 R$. The first condition is satisfied if

$$
\begin{gathered}
36 R^{2}(2 n+1)^{2}-48 R^{2}(4 n+1) \geqq 0 \\
12 R^{2}(2 n-1)(6 n+1) \geqq 0 \\
n \geqq \frac{1}{2}
\end{gathered}
$$

To determine whether one of the roots is between 0 and $2 R$, we substitute these values for $m$ in 1); the substitution values are

$$
f(0)=12 R^{2} \quad f(2 R)=4 R^{2}(1-2 n)
$$

The first is positive, the second is $\leqq 0$ according to 2), so the smaller of the roots of 1) always satisfies the condition if $n \geqq \frac{1}{2}$.

The problem was solved by: Baruch Jenö, fg. VIII. Nyíregyháza; Grossmann Gusztáv, fg. VIII. Budapest; ifj. Imre János, fg. VIII. Nyíregyháza; Jankovich György, fg. VIII. Losoncz; Visnya Aladár, fr. VII. Pécs.