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Problem 1123

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Geometry Difficulty 5.0 Prove it India — Team Selection Test · India · 2010

Let $ABC$ be a triangle in which $BC < AC$ . Let $M$ be the mid-point of $AB$ ; $AP$ be the altitude from $A$ on to $BC$ ; and $BQ$ be the altitude from $B$ on to $AC$ . Suppose $QP$ produced meets $AB$ (extended) in $T$ . If $H$ is the ortho-centre of $ABC$ , prove that $TH$ is perpendicular to $CM$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Complete the parallelogram $ADBC$ . Join $CD$ , $CH$ and $HD$ . Let $S$ and $L$ be the midpoints of $CH$ and $HD$ respectively. Observe that $CPHQ$ is a cyclic quadrilateral and $CH$ is a diameter of the circumscribing circle. Thus $S$ is the centre of a circle $\Gamma_1$ passing through $H, P, Q, C$ . Similarly, $L$ is the centre of a circle $\Gamma_2$ passing through $A, D, B, H$ . Hence the radical axis of these circles, which passes through $H$ is perpendicular to $SL$ . However $SL$ is parallel to $CD$ (which passes through $M$ ) and hence the radical axis of $\Gamma_1$ and $\Gamma_2$ is perpendicular to $CM$ . But $BPQA$ is also cyclic so that $TP \cdot TQ = TB \cdot TA$ . This shows that $T$ has same power with respect to $\Gamma_1$ and $\Gamma_2$ . Thus $T$ is on the radical axis of these circles. It follows that $TH$ is perpendicular to $CM$ .

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