Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Saudi Arabia

Let $ABC$ be an acute, non-isosceles triangle with $H$ is the orthocenter and $M$ is the midpoint of $AH$ . Denote $O_1, O_2$ as the centers of circles pass through $H$ and respectively tangent to $BC$ at $B, C$ . Let $X, Y$ be the ex-centers which respect to angle $H$ in triangles $HMO_1, HMO_2$ . Prove that $XY$ is parallel to $O_1O_2$ .

Solution

Let $BU$ , $CV$ be the diameter of circles $(O_1)$ , $(O_2)$ and denote $R = HU \cap AB$ , $S = HV \cap AC$ . We have $BH \perp HU$ , but $BH \perp AC$ then $HR \parallel AC$ . Similarly $HS \parallel AB$ implying that $ARHS$ is a parallelogram. Hence, $M$ is the midpoint of the segment $RS$ .

In the other hand, $O_1$ is the midpoint of $BU$ and $BU \parallel AH$ thus, by the property of trapezoid, we get $O_1, R, M$ are collinear. Similarly, $O_2, S, M$ are also collinear. Hence, five points $O_1, O_2, R, S, M$ are collinear.

Since $X$ is the excenter of triangle $HMO_1$ then $MX$ is the external angle bisector of $\angle O_1MH$ , thus $\frac{XR}{XH} = \frac{MR}{MH}$ . Similarly, $\frac{YS}{YH} = \frac{MS}{MH}$ , but $MR = MS$ then $\frac{XR}{XH} = \frac{YS}{YH}$ , this implies that $XY \parallel RS$ or $XY \parallel O_1O_2$ .

$\square$

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