29° Olimpiada Matemática del Cono Sur

By considering the inscribed angle $\angle BAC$ in the circumcircle of the triangle $ABC$, we have that $\angle BOC=2\cdot \angle BAC=120^{\circ }$; then, $\angle B{O}^{\prime }C=60^{\circ }$. Since $O$ is a point of the perpendicular bisector of $BC$, then $O^{\prime }$ is also on this line. Therefore, the triangle $B{O}^{\prime }C$ is equilateral.

On the other hand, we have that $\angle BIC=90^{\circ }+\frac{1}{2}\angle BAC=120^{\circ }$. It follows that $I$ is on the circumcircle of $BOC$. By applying Ptolemy's theorem to the quadrilateral $BIC{O}^{\prime }$, we obtain: $$I{O}^{\prime }\cdot BC=BI\cdot O^{\prime }C+CI\cdot O^{\prime }B$$ and, recalling that $BC=O^{\prime }C=O^{\prime }B$, we conclude that $$I{O}^{\prime }=BI+CI.$$