Given $AB = 2a$ is a line segment with midpoint $C$. We draw a line $XY$ through $C$ that makes a $60^{\circ}$ angle with $CB$. Choose a point $P$ on $XY$ such that the ratio of the lateral areas of the cones formed by the rotation of $PA$ and $PB$ around $XY$ is equal to a given value $m$. For which values of $m$ is this problem possible?
Let's raise perpendiculars from $X Y$ to $A A_{1}$ and $B B_{1}$. The resulting triangles $A C A_{1}$ and $B C B_{1}$ are congruent, hence $A A_{1}=B B_{1}$.

The lines $A P$ and $P B$ rotating around $X Y$ describe conical surfaces with areas $\pi A A_{1} \cdot A P$ and $\pi B B_{1} \cdot B P$, respectively, and the ratio of these areas is:

$$
\frac{A P}{B P}=m
$$

Let $C P=x$, then

$$
\begin{aligned}
& \overline{A P}^{2}=\left(x+A_{1} C\right)^{2}+a^{2}-{\overline{C A_{1}}}^{2} \\
& \overline{B P}^{2}=\left(x-B_{1} C\right)^{2}+a^{2}-{\overline{B_{1} C}}^{2}
\end{aligned}
$$

Since

$$
A_{1} C=B_{1} C=\frac{a}{2}
$$

it follows that

$$
\frac{\overline{A P}^{2}}{\overline{B P}^{2}}=\frac{a^{2}+x^{2}+a x}{a^{2}+x^{2}-a x}=m^{2}
$$

or

$$
\left(m^{2}-1\right) x^{2}-a\left(m^{2}+1\right) x+a^{2}\left(m^{2}-1\right)=0
$$

For the problem to be feasible, it is necessary and sufficient that the value of $x$ be real, that is:

$$
a^{2}\left(m^{2}+1\right)^{2}-4 a^{2}\left(m^{2}-1\right)^{2} \geqq 0
$$

or, expanding,

$$
\left(3 m^{2}-1\right)\left(3-m^{2}\right) \geqq 0
$$

The inequality holds if

$$
\frac{1}{3} \leqq m^{2} \leqq 3
$$

or

$$
\frac{\sqrt{3}}{3} \leqq m \leqq \sqrt{3}
$$

(Pál Sárközy, Pannonhalma.)