Given a circle with center $O$ and radius $r$. Write a variable triangle $ABC$ around this circle such that the product $OA \cdot OB \cdot OC = p^3$ is constant. Find the geometric locus of the center of the circumcircle of triangle $ABC$.
Let the radius of the inscribed circle be $r$, the radius of the circumscribed circle be $R$; the center of the circumscribed circle is $P$.

Since

$$
O A=\frac{r}{\sin \frac{A}{2}}, O B=\frac{r}{\sin \frac{B}{2}}, \text { and } O C=\frac{r}{\sin \frac{C}{2}}
$$

therefore

$$
p^{3}=O A \cdot O B \cdot O C=\frac{r^{3}}{\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}=\frac{r^{3}}{\frac{r}{4 R}}=4 R r^{2}
$$

from which

$$
R=\frac{p^{3}}{4 r^{2}}
$$

The distance between the centers of the two circles is

$$
O P=\sqrt{R^{2}-2 R r}=\frac{p}{4 r^{2}} \sqrt{p\left(p^{3}-8 r^{3}\right)}
$$

Thus, $O P$ is constant, and the geometric locus of point $P$ is a circle centered at $O$ with radius

$$
O P=\frac{p}{4 r^{2}} \sqrt{p\left(p^{3}-8 r^{3}\right)}
$$

The problem is solvable if $p \geq 2 r$; if $p=2 r$, and $O$ and $P$ coincide. It is evident that the problem of inscribing a triangle in one of two concentric circles and circumscribing it around the other is only solvable if the radius of the latter is twice that of the former. The triangle $A B C$ is then equilateral.

(Riesz Frigyes.)

The problem was also solved by: Friedmann B., Kertész L., Kornis Ö., Roth M., Sasvári G., Szabó K.