1st Iberoamerican 1985

The reals x, y, z satisfy x ≠ 1, y ≠ 1, x ≠ y, and (yz - x²)/(1 - x) = (xz - y²)/(1 - y). Show that (yx - x²)/(1 - x) = x + y + z.

Solution

We have yz - x² - y²z + yx² = xz - y² - x²z + xy². Hence z(y - x - y² + x²) = -y² + xy² - x²y + x². Hence z = (x + y - xy)/(x + y - 1).

So yz = x + y + z - xy - xz, so yz - x² = x + y + z - x² - xy - xz = (x + y + z)(1 - x), so (yz - x²)/(1 - x) = (x + y + z).

1st Ibero 1985

© John Scholes
jscholes@kalva.demon.co.uk
1 July 2002