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.

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TopNewestI 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. – Samuraiwarm Tsunayoshi · 2 years ago

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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. – Xuming Liang · 2 years ago

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