Show that if the real numbers x, y, z satisfy 1/(yz - x2) + 1/(zx - y2) + 1/(xy - z2) = 0, then x/(yz - x2)2 + y/(zx - y2)2 + z/(xy - z2)2 = 0.
Solution
Solution by Demetres Christofides
Put yz - x2 = k (1), zx - y2 = h (2). Then the given equality implies xy - z2 = -hk/(h + k). Evidently h and k are non-zero (or the given equality would not hold).
y (1) + z (2) + x(3) gives: ky + hz - hkx/(h + k) = 0. Hence hkx = hky + k2y + h2z + hkz. Similarly, z (1) + x (2) + y (3) gives: hky = h2x + hkx + hkz + k2z. Adding, we get 0 = h2x + k2y + (h + k)2z. Dividing by h2k2 gives the required result.
© John Scholes
jscholes@kalva.demon.co.uk
11 Sep 2002