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 AB2 + BC2 + CA2.
(ii) Find the locus of the midpoint of BC.
Solution
(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 AP2 = 4r2 - 4x2. BM2 = R2 - OM2 = R2 - (r2 - x2). That is essentially all we need, because we now have: AB2 + AC2 + BC2 = (AP2 + (BM - x)2) + (AP2 + (BM + x)2) + 4BM2 = 2AP2 + 6BM2 + 2x2 = 2(4r2 - 4x2) + 6(R2 - r2 + x2) + 2x2 = 6R2 + 2r2 , 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.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
25 Oct 1998