The 16th Japanese Mathematical Olympiad - The First Round

Let $P$, $Q$ and $R$ be the circumcenters of the triangles $OBC$, $OCA$ and $OAB$, respectively, and let $r$ be their common circumradius.

Since $PB=PO=RB=RO=r$, the quadrilateral $BPOR$ is a rhombus and so $\overrightarrow{BR}=\overrightarrow{PO}$. Similarly, $\overrightarrow{CQ}=\overrightarrow{PO}$. So $\overrightarrow{BR}=\overrightarrow{CQ}$, and it follows that $BCQR$ is a parallelogram. Therefore $BC=QR$.

Similarly $CA=RP$ and $AB=PQ$ follows, and we obtain $△ABC\cong △PQR$. Since the points $P$, $Q$ and $R$ are on the circle with center $O$ and radius $r$, the circumradius of triangle $PQR$ is $r$. So $r$ is equal to the circumradius of triangle $ABC$.

By cosine formula, $\cos \angle B=\frac{A{B}^2+B{C}^2-C{A}^2}{2AB\cdot BC}=\frac{1}{2}$. By sine formula, $r=\frac{CA}{2\sin \angle B}=\frac{7}{\sqrt{3}}$.