A right circular cone has a height of $m$, and the radius of its base is $r$; let's calculate the radius of the sphere that can be inscribed in the cone.
Cut the cone with a plane passing through its axis. In this case, the resulting section will be an isosceles triangle $ABC$, whose base is $BC=2r$, height is $AM=m$, and the radius of the inscribed circle is $OD=x$, which is the radius of the sought sphere. Since $\triangle ABM \sim \triangle ADO$, we have

$$
BM : AM = DO : AD
$$

or

$$
r : m = x : \sqrt{r^2 + m^2} - r
$$

from which

$$
x = \frac{\sqrt{r^2 + m^2} - r}{m} r
$$

(Lajos Neumann, Budapest, V. district.)

The problem was also solved by: Babocsai Gy., Bayer N., Czukor G., Erdős V., Fried E., Grossberger Z., Havas B., Herskovics S., Kirchknopf E., Koffler B., Kovács Gy., Paunz A., Sárközy P., Vilcsek A., Weisz A.