13th CanMO 1981 Problem 2 The circle C has radius 1 and touches the line L at P. The point X lies on C and Y is the foot of the perpendicular from X to L. Find the maximum possible value of area PXY (as X varies). Solution
: (3√3)/8. Let O be the center of C and QP a diameter. Let ∠QOX = x. Then PY = sin x and XY = 1 + cos x, so the area is sin x(1 + cos x)/2. We claim that this has minimum value (3√3)/8. Put y = sin x, so √(1 - y 2 ) = cos x. We have (y - 3√3/4y) 2 ≥ 0, so y 2 - (3√3)/2y + 27/16y 2 ≥ 0, or 27/16y 2 - (3√3)/2y + 1 ≥ 1 - y 2 (*). Now y ≤ 1, so (3√3)/4y > 1. Thus taking the positive square roots of (*) we get: (3√3)/4y - 1 ≥ √(1 - y 2 ) or y(1 + √(1 - y 2 )) ≤ (3√3)/4, as claimed. 13th CanMO 1981 © John Scholes jscholes@kalva.demon.co.uk 21 June 2002 Last corrected/updated 20 Oct 03