Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Greece

Let $AB\Gamma$ be an acute angled triangle with $AB < A\Gamma$ and circumcenter $O$ . The altitudes $B\Delta$ , $\Gamma E$ meet at point $H$ . If $O_1$ is the circumcenter of the triangle $BH\Gamma$ , prove that the quadrilateral $AHO_1O$ is parallelogram.

Solution

Since $O_1$ belongs to perpendicular bisector $OM$ of the segment $B\Gamma$ and $AH$ , $OO_1$ , it is enough to prove that $AH = OO_1$ . Since $AH = 2OM$ , it is enough to prove that $OM = MO_1$ . The quadrilateral is cyclic, whereby $\hat{BH}\Gamma = 180^\circ - \hat{A}$ . Moreover $\hat{BO}_1\Gamma = 2\hat{A}$ . Therefore the isosceles triangles $BO\Gamma$ , $BO_1\Gamma$ have all their corresponding angles equal and since they have $B\Gamma$ common side, they are equal. Therefore $B\Gamma$ is the perpendicular bisector of $OO_1$ , whereby $M$ is the midpoint of $O_1O$ , and hence $OM = MO_1$ .

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