Japan Mathematical Olympiad

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 $$\frac{r_1}{r_2}=\frac{PQ}{QR}=\frac{XQ}{XR}=\sqrt[3]{\frac{2}{9}}.$$