For a certain isosceles triangle, the volume $v$ generated by its rotation around its height is known, as well as the radius $\rho$ of the inscribed circle. Find the height and base of the triangle.
Let the base of the isosceles triangle be $2a$, the height $m$, and the side $b$. Then, from the formula for the volume of a cone and the radius of the inscribed circle of the triangle, the following equations can be written:

$$
\begin{gathered}
\frac{3 v}{\pi}=a^{2} m \\
p=\frac{2 t}{2 a+2 b}=\frac{t}{a+b}=\frac{a m}{a+\sqrt{a^{2}+m^{2}}}
\end{gathered}
$$

From (2),

$$
a^{2}=\frac{m \rho^{2}}{m-2 \rho}
$$

From (1),

$$
a^{2}=\frac{3 v}{\pi m}
$$

Equating the two values and rearranging the equation:

$$
\rho^{2} \pi m^{2}-3 v m+6 v \rho=0
$$

From which,

$$
m=\frac{3 v \pm \sqrt{9 v^{2}-24 \rho^{3} v \pi}}{2 \rho^{2} \pi}
$$

Substituting this value into (1):

$$
2 a=\frac{3 v \mp \sqrt{9 v^{2}-24 \rho^{3} v \pi}}{\rho \pi}
$$

The problem is possible if

$$
9 v^{2}>24 \rho^{3} v \pi
$$

or

$$
v>\frac{8 \rho^{3} \pi}{3}
$$

(Mór Klein, Győr.)

The problem was also solved by Bálint B., Bauss O., Devecis M., Freund A., Friedmann B., Geist E., Goldstein Zs., Grünhut B., Hofbauer E., Iványi B., Kántor N., Ländler D., Orlowszky F., Riesz F., Schiffer H., Spitzer Ö., Szabó I., Szabó K., Szilágyi Z.