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IMO 2015 Problem 3

Let $ABC$ be an acute triangle with $AB\gt AC$. Let $\Gamma$ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HQA=90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ=90^{\circ}$. Assume that the points $A,B,C,K$ and $Q$ are all different and lie on $\Gamma$ in this order. Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.


I don't want to find a full answer. Instead, I only want to show $ Q, H, M $ are collinear.

Let $AE$ be the diameter of $\Gamma$, then $BECH$ is a parallelogram. then, $ H, M , E$ are collinear. $HQ\perp AQ \implies$ $Q, H, M , E$ are collinear.

If one don't prove that $BECH$ is a parallelogram, Is there any other way to show $Q, H, M $ are collinear? Any help will be appreciated!

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  • $\begingroup$ Def. of $\Gamma$? $\endgroup$ – aGer Jul 16 '15 at 21:55
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Since $OM \perp BC$, we have $OM= \dfrac{AH}{2}$. Draw the line $EH$. Also, draw $OM' \parallel AH $ to meet $EH$ at $M'$. Consequently, $OM'=\dfrac{AH}{2}$, and $OM' \perp BC \implies OM=OM'$.

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  • $\begingroup$ who is the point M'? $\endgroup$ – user85046 Jul 17 '15 at 5:22
  • $\begingroup$ @CFG,add the details. $\endgroup$ – chenbai Jul 17 '15 at 9:22

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