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.

**Solution**

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.

© John Scholes

jscholes@kalva.demon.co.uk

1 Jan 2001