A triangle has sides $a=10 \mathrm{~cm}, b=15 \mathrm{~cm}, c=20 \mathrm{~cm}$. What is the volume of the double cone that is formed when the triangle rotates around its longest side?
Given the height $m$ corresponding to side $c$, the volume of the double cone is

$$
K=\frac{m^{2} \pi c}{3}
$$

Let $a$ be the angle opposite to side $a$. According to Carnot's theorem,

$$
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha
$$

from which

$$
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}
$$

or

$$
\sin ^{2} \alpha=1-\cos ^{2} \alpha=\frac{4 b^{2} c^{2}-\left(b^{2}+c^{2}-a^{2}\right)^{2}}{4 b^{2} c^{2}}
$$

But

$$
m^{2}=b^{2} \sin ^{2} \alpha=\frac{4 b^{2} c^{2}-\left(b^{2}+c^{2}-a^{2}\right)^{2}}{4 c^{2}}
$$

thus

$$
K=\frac{4 b^{2} c^{2}-\left(b^{2}+c^{2}-a^{2}\right)^{2}}{12 c} \pi
$$

Substituting the given values, we obtain

$$
K=1103.904 \mathrm{~cm}^{3}
$$

(Pál Sárközy, Pannonhalma.)

The problem was also solved by: Babocsai Gy., Chambrée M., Czukor G., Erdős V., Fried E., Friedrich P., Kirchknopf E., Koffler B., Kovács Gy., Miklóssy K., Neumann L., Paunz A., Rajz E., Tóth B., Viola R.