IMO 1988 Problem A1 Consider two coplanar circles of radii R > r with the same center. Let P be a fixed point on the smaller circle and B a variable point on the larger circle. The line BP meets the larger circle again at C. The perpendicular to BP at P meets the smaller circle again at A (if it is tangent to the circle at P, then A = P). (i) Find the set of values of AB 2 + BC 2 + CA 2 . (ii) Find the locus of the midpoint of BC.
(i) Let M be the midpoint of BC. Let PM = x. Let BC meet the small circle again at Q. Let O be the center of the circles. Since angle APQ = 90 degrees, AQ is a diameter of the small circle, so its length is 2r. Hence AP 2 = 4r 2 - 4x 2 . BM 2 = R 2 - OM 2 = R 2 - (r 2 - x 2 ). That is essentially all we need, because we now have: AB 2 + AC 2 + BC 2 = (AP 2 + (BM - x) 2 ) + (AP 2 + (BM + x) 2 ) + 4BM 2 = 2AP 2 + 6BM 2 + 2x 2 = 2(4r 2 - 4x 2 ) + 6(R 2 - r 2 + x 2 ) + 2x 2 = 6R 2 + 2r 2 , which is independent of x. (ii) M is the midpoint of BC and PQ since the circles have a common center. If we shrink the small circle by a factor 2 with P as center, then Q moves to M, and hence the locus of M is the circle diameter OP. s are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X. 29th IMO 1988 © John Scholes jscholes@kalva.demon.co.uk 25 Oct 1998