10th CanMO 1978 Problem 3 Find the real solution x, y, z to x + y + z = 5, xy + yz + zx = 3 with the largest z.
We have 25 = (x + y + z) 2 = (x 2 + y 2 + z 2 ) + 2(xy + yz + zx) = (x 2 + y 2 + z 2 ) + 6, so (x 2 + y 2 + z 2 ) = 19. Hence (x - 1/3) 2 + (y - 1/3) 2 + (z - 1/3) 2 = (x 2 + y 2 + z 2 ) -2(x + y + z)/3 + 1/3 = 19 - 10/3 + 1/3 = 16. So the maximum possible value of z is 4 1/3 (with solution x = 1/3, y = 1/3, z = 4 1/3). 10th CanMO 1978 © John Scholes jscholes@kalva.demon.co.uk 10 June 2002