IMO 1991 SL 1

Origin: PHI

Problem

Let ABC be any triangle and P any point in its interior. Let P1, P2 be the feet of the perpendiculars from P to the two sides AC and BC. Draw AP and BP, and from C drop perpendiculars to AP and BP. Let Q1 and Q2 be the feet of these perpendiculars. Prove that the lines Q1P2, Q2P1, and AB are concurrent.

Solution

All the angles \anglePP1C, \anglePP2C, \anglePQ1C, \anglePQ2C are right, hence P1, P2, Q1, Q2 lie on the circle with di- ameter PC. The result now fol- lows immediately from Pascal’s theorem applied to the hexagon P1PP2Q1CQ2. It tells us that the points of intersection of the three pairs of lines P1C, PQ1 (intersec- tion A), P1Q2, P2Q1 (intersection A B C P P2 P1 Q1 Q2 X X) and PQ2, P2C (intersection B) are collinear.