IMO Shortlist 1991 problem 2


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2. travnja 2012.
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ABC is an acute-angled triangle. M is the midpoint of BC and P is the point on AM such that MB = MP. H is the foot of the perpendicular from P to BC. The lines through H perpendicular to PB, PC meet AB, AC respectively at Q, R. Show that BC is tangent to the circle through Q, H, R at H.

Original Formulation:

For an acute triangle ABC, M is the midpoint of the segment BC, P is a point on the segment AM such that PM = BM, H is the foot of the perpendicular line from P to BC, Q is the point of intersection of segment AB and the line passing through H that is perpendicular to PB, and finally, R is the point of intersection of the segment AC and the line passing through H that is perpendicular to PC. Show that the circumcircle of QHR is tangent to the side BC at point H.
Izvor: Međunarodna matematička olimpijada, shortlist 1991