THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Let $A_1$ be the intersection point of the perpendicular bisectors of the line segments $IB$ and $IC$. Let the angle bisector $AI$ intersect the circumcircle of $ABC$ at $D$. Since $\angle BID=\angle DBI$ and $\angle CID=\angle DCI$, it follows that $DB=DI=DC$, hence $A_1=D$. Thus, $A_1$ belongs to the circumcircle of $ABC$ and the same goes for $B_1$ and $C_1$.

Observe that $I$ is the orthocenter of $\overline{A_1{B}_1{C}_1}$ and since $O$ is its circumcenter, Sylvester's relation yields $OI=O{A}_1+O{B}_1+O{C}_1$, as desired.