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# Problem 3! IMO 2015

Let $$ABC$$ be an acute triangle with $$AB > AC$$. Let $$\Gamma$$ be its cirumcircle, $$H$$ 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.

##### This is part of the set IMO 2015

Note by Sualeh Asif
1 year, 10 months ago

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I don't know if that works or not, but try to use Nine-point circle to prove it. And I think that the point $$K$$ is probably Feuerbach point of some triangle. · 1 year, 10 months ago

I can imagine that inversion kills this problem, here's my solution without using that, I don't expect many to understand this at this point as I wrote it in a hurry.

Let $$I$$ denote the midpoint of $$HQ$$. It suffices to show that the circumcenter of $$KFM$$ lies on $$IK$$, which equates to proving $$\angle IKM=\angle 90-\angle KFC=\angle KFH$$.

$$AH,QH,KH$$ are extended to meet $$(O)$$ again at $$E,Z,X$$ respectively. It is well known and obvious that $$O\in QX, AZ, M\in QZ$$. Let $$F'$$ denote the midpoint of $$XH$$, $$AF'\cap (O)=Y$$.

Since $$(XZ\parallel MF')\perp (HQ$$, therefore $$XZ\parallel MF'\parallel AQ$$. By the converse of Pascal's theorem, we can deduce that $$AF'\cap KM=Y$$. Note that $$\triangle AHXF'\sim \triangle KHEF$$, thus $$\angle FKE=\angle XAF'=\angle XKY$$. Additionally, $$\angle IKH=\angle OKA=\angle AZK=\angle AEK$$ by spiral similarity about $$K$$ between $$(I),(O)$$. Hence $$\angle HFK=\angle AEK+\angle FKE=\angle IKH+\angle XKY=\angle IKM$$ and we are done. · 1 year, 10 months ago