32nd IMO shortlist 1991/1 solution

32nd IMO 1991 shortlist

Problem 1

ABC is a triangle and P an interior point. Let the feet of the perpendiculars from P to AC, BC be P₁, P₂ respectively, and let the feet of the perpendiculars from C to AP, BP be Q₁, Q₂ respectively. Show that P₁Q₂ and P₂Q₁ meet on the line AB.

Solution

All the angles PP₁C, PP₂C, PQ₁C, PQ₂C are 90^o, so P₁, P₂, Q₁, Q₂ lie on the circle diameter PC.

The result now follows 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 (intersection A), and P₁Q₂, P₂Q₁ (intersection X), and PQ₂, P₂C (intersection B) are collinear.

32nd IMO shortlist 1991