# Other way to show $Q, H, M$ are collinear without $BECH$ is a parallelogram (IMO 2015 Problem 3 )

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!

• Def. of $\Gamma$? – aGer Jul 16 '15 at 21:55

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 $$OM' \perp BC \implies OM=OM'$$.