The tricky part is deciding what to use as variables. Take x, y coordinates centered on the center of the circle. Assume that the diagonals intersect at (0, h) and that one of the diagonals has gradient m, so that it has equation y = mx + h. Then it cuts the circle at x-coordinates x 1 , x 2 which are the roots of the quadratic: x 2 + (mx + h) 2 = 1, or (m 2 + 1)x 2 + 2mhx - (1 - h 2 ) = 0. Since one root is positive and the other negative we have that d = 2 x sum of roots = 4mh/(m 2 + 1). So d/h = 2. 2m/(m 2 + 1). But (m - 1) 2 ≥ 0 implies 2m/(m 2 + 1) ≥ 1, with equality iff m = 1. Hence sup d/h = 2, with equality when the diagonals are perpendicular. 50th Putnam 1989 © John Scholes jscholes@kalva.demon.co.uk 1 Jan 2001
50th Putnam 1989 Problem B5 A quadrilateral is inscribed in a circle radius 1. Two opposite sides are parallel. The difference between their lengths is d > 0. The distance from the intersection of the diagonals to the center of the circle is h. Find sup d/h and describe the cases in which it is attained.