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Iranian Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Iran

Let $ABC$ be an acute-angled triangle. The altitudes $BE$ , $CF$ meet at $H$ . $O$ is the circumcenter of triangle $ABC$ . $M$ is the midpoint of $BC$ . $P$ is the point on $EF$ such that $HP \perp HO$ . $Q$ is the point on $HA$ such that $PQ \perp HM$ . Prove that $QA = 3QH$ .

Solution

Let $(O)$ be the circumcircle of triangle $ABC$ . $AK$ is the diameter of $(O)$ . If $D$ , $S$ , $T$ are the reflection points of $H$ through $BC$ , $CA$ , $AB$ , then $D$ , $S$ , $T$ are on $(O)$ . It's easily seen that $HCKB$ is a parallelogram so $M$ lies on $HK$ . Let $KH$ intersect $(O)$ again at $G$ . Lines $HP$ intersects the lines $BC$ , $DK$ , $ST$ , $GA$ at $R$ , $X$ , $Y$ , $Z$ respectively.

Note that $EF$ is the median line of triangle $HST$ so $HY = 2HP$ . Because $OH \perp YR$ , according to butterfly theorem for segments $YR$ and $ZX$ , it is obtained for quadrilateral $BCTS$ that $HR = HY = 2HP$ . By butterfly theorem for quadrilateral $AGDK$ it is also obtained that $HZ = HX = 2HR = 2HY = 4HP$ . Note that $PQ \perp HM$ so $PQ \parallel AG$ . Noticing the parallel lines, it is deduced that $HA = 4HQ$ , this means $QA = 3QH$ . Hence the proof is done. ■

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