Iranian Mathematical Olympiad

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. ■