-abc/(ab+bc+ca) Solution Label the equations (0),(1),(2),(3). Suppose c = 0. Then a, b ≠ 0. Note also that (a-b), (b-c) and (c-a) ≠ 0. (1) and (2) give x = w 2 (w-b)/(a 2 (a-b)), y = w 2 (a-w)/(b 2 (a-b)). Substituting in (3) we get w = 0 or, after some reduction w 2 - (a+b)w + ab = 0, so w = a or b. But w, a, b, c are all distinct, so we must have w = 0. Checking, we see that x = y = 0, z = 1, w = 0 certainly satisfies the equations if c = 0. Similarly, if a or b = 0, then w must be 0. So suppose a, b, c are all non-zero. Taking a(1)-(2), we get yb 2 (a-b) + zc 2 (a-c) = w 2 (a-w). Taking a(2)-(3), we get yb 3 (a-b) + zc 3 (a-c) = w 3 (a-w). Hence y = w 2 (a-w)(c-w)/(b 2 (a-b)(c-b)). Similarly, x = w 2 (b-w)(c- w)/(a 2 (b-a)(c-a)) and z = w 2 (a-w)(b-w)/(c 2 (a-c)(b-c)). Now substituting in (0) gives a quartic in w which has no w term. But by inspection we see that there is a solution x = 1, y = z = 0 for w = a, so w = a must be one root of the quartic. Similarly, b and c must be roots. Suppose the other root is k. Then since the coefficient of w is 0, we have abc + abk + bck + cak = 0. Hence k = -abc/(ab+bc+ca). Note that this also gives the correct result if any of a, b, c are 0. 7th Irish 1994 © John Scholes [email protected] 2 February 2004 Last updated/corrected 2 Feb 04