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Romanian Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it Romania

In the triangle $ABC$ we denote by $O$ and $I$ the circumcenter and the incenter, respectively. The perpendicular bisectors of the line segments $IA$ , $IB$ and $IC$ pairwise intersect, thus defining the triangle $A_1B_1C_1$ . Prove that
$\vec{OI} = \vec{OA_1} + \vec{OB_1} + \vec{OC_1}.$

Solution

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 $\overline{A_1B_1C_1}$ and since $O$ is its circumcenter, Sylvester's relation yields $OI = OA_1 + OB_1 + OC_1$ , as desired.

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