We inscribe an equilateral cylinder in an equilateral cone. How does the volume of the cone compare to the volume of the cylinder?
The axis section of an equilateral cone is an equilateral triangle, and that of an equilateral cylinder is a square. Let $a$ be the slant height of the cone, $x$ the height of the cylinder, and $m$ the height of the cone. From similar triangles:

$$
a: x=m: m-x
$$

from which

$$
x=\frac{a m}{a+m}
$$

and since

$$
m=\frac{a}{2} \sqrt{3}
$$

therefore

$$
x=\frac{a \sqrt{3}}{2+\sqrt{3}} .
$$

If $v_{k}$ is the volume of the cone, and $v_{h}$ that of the cylinder, then:

$$
\begin{gathered}
v_{k}=\frac{1}{3} \frac{a^{2} \pi}{4} \cdot \frac{a}{2} \sqrt{3}=\frac{a^{3} \pi}{24} \sqrt{3} \\
v_{h}=\frac{x^{2}}{4} \pi \cdot x=\frac{x^{3} \pi}{4}=\frac{3 a^{3} \pi}{4(2+\sqrt{3})^{3}} \sqrt{3}
\end{gathered}
$$

and thus

$$
v_{k}: v_{h}=\frac{a^{3} \pi}{24} \sqrt{3}: \frac{3 a^{3} \pi}{4(2+\sqrt{3})^{3}} \sqrt{3}
$$

or

$$
\begin{gathered}
v_{k}: v_{h}=(2+\sqrt{3})^{3}: 18 \\
v_{k}: v_{h}=(26+15 \sqrt{3}): 18
\end{gathered}
$$

(Pál Pilczer, Kaposvár.)

The problem was also solved by: F. Aczél, I. Bartók, B. Bayer, G. Bogdán, Gy. Hirschfeld, F. Kertész, A. Klein, D. König, L. Lázár, Gy. Lukhaub, G. Messik, Gy. Póka, J. Sasvári, J. Scharff, V. Spitzer, M. Steiner, Gy. Sümegi, F. Winter, S. Wohlstein.