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Silk Road Mathematics Competition Xvii

Geometry Difficulty 5.9 AIME, harder Prove it Silk Road Mathematics Competition

In the triangle $ABC$ , points $H, L, K$ are chosen on the sides $AB, BC, AC$ , respectively, so that $CH \perp AB$ , $HL \parallel AC$ , $HK \parallel BC$ . In the triangle $HBL$ , let $P$ and $Q$ be feets of altitudes from $H$ and $B$ , respectively. Prove that in the triangle $AKH$ , the feets of altitudes from $A$ and $H$ lie on the line $PQ$ .

Solution

Suppose $PQ$ intersects $AK$ and $HK$ at $R$ and $S$ , respectively. From the fact that $P$ and $Q$ lie on a circle with diameter $HB$ and from $HL \parallel AC$ , it follows that $\angle LPQ = \angle BHQ = \angle BAR$ , i.e. the points $A, R, P$ and $B$ lie on the same circle. Therefore, $CP \cdot CB = CR \cdot CA$ . Also from the right triangle $BCH$ , we have $CH^2 = CP \cdot CB$ , implying $CH^2 = CR \cdot CA$ and hence $HR \perp AC$ .

It remains to prove that $AS \perp HK$ . From the fact that $A, R, P$ and $B$ lie on one circle and $HK \parallel BC$ it follows that $\angle CRP = \angle CBA = \angle KHA$ , i.e. the points $A, R, S$ and $H$ lie on the same circle. Since $\angle ARH = 90^\circ$ , we have $\angle ASH = 90^\circ$ and $AS \perp HK$ .

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