The sides of a triangle are given. Let's calculate the volume of the solid of revolution formed by rotating the triangle around one of its sides. Around which side should the triangle be rotated so that the volume of the resulting solid is maximized?
The volume of a solid of revolution is the sum or difference of the volumes of two cones with a common base. Let the height corresponding to the side $a$ of the triangle be $AD = m_1$, and let $BD = x$ and $DC = a - x$. If the triangle rotates around the side $a$, the volume of the resulting solid of revolution is:

$$
V_{1}=\frac{m_{1}^{2} \pi x}{3}+\frac{m_{1}^{2} \pi(a-x)}{3}=\frac{a m_{1}^{2} \pi}{3}
$$

but

$$
a m_{1}=2 \sqrt{s(s-a)(s-b)(s-c)}
$$

so

$$
V_{1}=\frac{4 \pi}{3 a} s(s-a)(s-b)(s-c)
$$

If the triangle rotates around the side $b$ or $c$, we similarly obtain that

$$
V_{2}=\frac{4 \pi}{3 b} s(s-a)(s-b)(s-c)
$$

and

$$
V_{3}=\frac{4 \pi}{3 c} s(s-a)(s-b)(s-c)
$$

We see that the triangle should rotate around its smallest side to maximize the volume of the resulting solid.

(Frigyes Orlowszky, Szamosujvár.)

The problem was also solved by: Beck F., Dénes A., Devecis M., Erdős A., Fekete J., Fleischmann S., Freibauer E., Goldziher K., Horovitz E., Kármán T., Kertész L., Posgay B., Spitzer Ö., Szabó I., Szabó K., Weisz J.