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Rushil
Navier-Stokes Equations

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Joined: 25 Jun 2005
Posts: 1608

One on quads
Indian IMOTC 2005 Day 4 Problem 1

Let $ABCD$ be a convex quadrilateral. The lines parallel to $AD$ and $CD$ through the orthocentre $H$ of $ABC$ intersect $AB$ and $BC$ Crespectively at $P$ and $Q$ . prove that the perpendicular through $H$ to th eline $PQ$ passes through th eorthocentre of triangle $ACD$

Posted: Fri Sep 23, 2005 8:09 am
Last edited by Rushil on Sat Oct 15, 2005 3:57 pm; edited 1 time in total

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darij grinberg
Birch & Swinnerton Dyer

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Joined: 10 Feb 2004
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Location: Karlsruhe / Munich

Re: Indian IMOTC 15

Problem. Let ABCD be a quadrilateral, and H the orthocenter of triangle ABC. The parallels to the lines AD and CD through the point H meet the lines AB and BC at the points P and Q, respectively. Prove that the perpendicular to the line PQ through the point H passes through the orthocenter of triangle ACD.

Solution. I will use the orthologic triangles theorem (quoted in http://www.mathlinks.ro/Forum/viewtopic.php?t=6472 post #5 and discussed as a warm-up exercise in http://www.mathlinks.ro/Forum/viewtopic.php?t=4337 )

Let G be the orthocenter of triangle ACD. Then, $AG\perp CD$ . On the other hand, HQ || CD. Thus, $AG\perp HQ$ , so that the line HQ is the perpendicular to the line AG through the point Q. Similarly, the line HP is the perpendicular to the line CG through the point P. Finally, since H is the orthocenter of triangle ABC, the line BH is the B-altitude of this triangle, i. e. it is the perpendicular to the line CA through the point B. Thus, the perpendiculars from the vertices B, P, Q of the triangle BPQ to the sidelines CA, CG, AG of the triangle GAC concur (namely, these perpendiculars are the lines BH, HP, HQ and thus concur at the point H). Hence, by the orthologic triangles theorem, it follows that the perpendiculars from the vertices G, A, C of the triangle GAC to the sidelines PQ, BQ, BP of the triangle BPQ concur. Now, the perpendicular from the point A to the line BQ is simply the perpendicular from the point A to the line BC, i. e. the A-altitude of triangle ABC, and similarly, the perpendicular from the point C to the line BP is the C-altitude of triangle ABC; hence, the point of concurrence, lying on both of these perpendiculars, must be the point of intersection of the A-altitude and the C-altitude of triangle ABC, i. e. simply the orthocenter H of triangle ABC. We thus obtain that the point H lies on the perpendicular from the point G to the line PQ. In other words, $GH\perp PQ$ . Thus, the perpendicular to the line PQ through the point H passes through the point G, i. e. through the orthocenter of triangle ACD. And the problem is solved.

darij

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Posted: Mon Sep 26, 2005 3:19 am

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