A right circular cylinder and a symmetrical double cone have the same height, the same surface area, and the same volume. Express their base radii, surface areas, and volumes in terms of their height.
A double cone refers to the surface described by a straight line when it rotates around another intersecting (but not perpendicular) straight line, thus the two cones in the ordinary sense share a common apex. Due to symmetry, the two base planes intersecting the surface are at equal distances from the common apex.

![](https://cdn.mathpix.com/cropped/2024_05_02_e12badce4527c65b6eadg-1.jpg?height=626&width=1206&top_left_y=294&top_left_x=449)

Let the radius of the cylinder's base be \( r \), the radius of the cone's base be \( R \), their common height be \( m \), their volume be \( V \), and their surface area be \( F \). Thus, the slant height of the two cones is the same, with length \( o = \sqrt{R^2 + \frac{m^2}{4}} \). From the equality of volumes,

\[
r^2 \pi m = 2 \frac{R^2 \pi m}{6}, \quad R^2 = 3 r^2
\]

(since clearly \( m \neq 0 \)). From the equality of surface areas,

\[
F = 2 r^2 \pi + 2 r \pi m = 2 \left( R^2 \pi + R \pi o \right) = 2 R^2 \pi + R \pi \sqrt{4 R^2 + m^2}
\]

Eliminating \( R \), simplifying, and squaring, we discard the negative root of the resulting quadratic equation:

\[
\begin{aligned}
& 2 m - 4 r = \sqrt{36 r^2 + 3 m^2} \\
& r = m \frac{\sqrt{21} - 4}{10} \approx 0.0583 \, \text{m}
\end{aligned}
\]

From this,

\[
\begin{gathered}
R = m \frac{3 \sqrt{7} - 4 \sqrt{3}}{10} \approx 0.1009 m, \quad V = \frac{m^3 \pi}{100} (37 - 8 \sqrt{21}) \approx 0.01055 m^3 \\
F = \frac{m^2 \pi}{50} (2 \sqrt{21} - 3) \approx 0.387 m^2
\end{gathered}
\]

Ferenc Naszalyi (Budapest, Kölcsey F. g. IV. o. t.)

Remark. Several people, starting from the visualization of problem 292 in physics, thought of the two cones joined at their bases. In this case, the term \( 2 R^2 \pi \) on the right side of (1) is missing, and similar calculations yield the following results:

\[
\begin{gathered}
r = m \frac{\sqrt{3} + 1}{8} \approx 0.342 m, \quad R = \frac{3 + \sqrt{3}}{8} m \approx 0.717 m \\
V = \frac{m^3 \pi}{32} (2 + \sqrt{3}) \approx 0.365 m^3, \quad F = \frac{m^2 \pi}{16} (6 + 5 \sqrt{3}) \approx 2.88 m^2
\end{gathered}
\]