Problem 2018

In view of a well-known theorem on angles subtended by arcs on a circle, we have $\angle PSQ = \angle QPR$, and $\angle SQP = \angle PRQ$. This implies that the triangles $PSQ$ and $QPR$ are similar triangles. Since the circles $O_1$ and $O_2$ are circum-circles of the triangles $PSQ$ and $QPR$, respectively, the ratio $\frac{r_1}{r_2}$ of the radii of these circles must be the same as the similarity ratio $\frac{PQ}{QR}$ of these triangles. The same theorem quoted above also tells us that we have $\angle XPS = \angle XQP = \angle XRQ$. Since the angle $\angle X$ is common to all of the three triangles $XPS$, $XQP$ and $XRQ$, we conclude that these triangles are similar to each other. Hence, we obtain $\frac{XS}{XP} = \frac{XP}{XQ} = \frac{XQ}{XR}$. Consequently, we get
$$\left(\frac{XQ}{XR}\right)^3 = \frac{XQ}{XR} \cdot \frac{XP}{XQ} \cdot \frac{XS}{XP} = \frac{XS}{XR} = \frac{2}{9}.$$
Finally, from the similarity of the triangles $XQP$ and $XRQ$, we also get $\frac{PQ}{QR} = \frac{XQ}{XR}$, which enables us to conclude that we have
$$\frac{r_1}{r_2} = \frac{PQ}{QR} = \frac{XQ}{XR} = \sqrt[3]{\frac{2}{9}}.$$