1st Balkan 1984 Problem 4 Given positive reals a, b, c find all real solutions (x, y, z) to the equations ax + by = (x - y) 2 , by + cz = (y - z) 2 , cz + ax = (z - x) 2 . Solution
: (x, y, z) = (0, 0, 0), (a, 0, 0), (0, b, 0), (0, c, 0). We have 2ax = (ax + by) - (by + cz) + (cz + ax) = 2(x 2 - xy + yz - xz) = 2(y - x)(z - x). Similarly, 2by = (ax + by) + (by + cz) - (cz + ax) = 2(y 2 + xz - xy - yz) = (z - y)(x - y) and 2cz = -(ax + by) + (by + cz) + (cz + ax) = 2(z 2 + xy - yz - zx) = 2(x - z)(y - z). Suppose z ≥ x, y. Then if x >= y, ax = (z - x)(y - x) ≤ 0 , and by = (z - y)(x - y) ≥ 0. But a and are positive, so x ≤ 0 and y ≥ 0. Hence x = y = 0. On the other hand, if x ≤ y, then ax ≥ 0 and by ≤ 0, so again x = y = 0. Hence cz = z 2 , so z = 0 or c. This gives the two solutions (x, y, z) = (0, 0, 0) and (0, 0, c). Similarly, if y ≥ x, z, then we find x = z = 0 and hence y = 0 or b. If x ≥ y, z, then we find y = z = 0 and x = 0 or a. 1st Balkan 1984 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002