3rd Brazil 1981 Problem 1 For which k does the system x 2 - y 2 = 0, (x-k) 2 + y 2 = 1 have exactly (1) two, (2) three real solutions?
(1) k = ±√2, (2) k = ±1. Solution We have (x-k) 2 + x 2 = 1, so 2x 2 - 2kx + k 2 -1 = 0. This has 0, 1 or 2 real solutions according as k 2 >2, =2 or <2. k = √2 gives x = 1/√2, y = 1/√2 or -1/√2, so there are two solutions to the original set. Similarly for k = -√2. If |k| < √2, then x = ½k ±½√(2-k 2 ), y = ± x. That gives 4 solutions unless one of the values of x is 0, in which case we get 3 solutions. So k = 1 gives x, y = 0, 0 or 1, 1 or 1, -1. k = -1 gives x, y = 0, 0 or -1, 1 or -1, -1. 3rd Brazil 1981 © John Scholes jscholes@kalva.demon.co.uk 23 Oct 2003 Last corrected/updated 23 Oct 03