17th VMO 1979

Show that for all x > 1 there is a triangle with sides, x⁴ + x³ + 2x² + x + 1, 2x³ + x² + 2x + 1, x⁴ - 1.

Solution

Put a = x⁴ + x³ + 2x² + x + 1, b = 2x³ + x² + 2x + 1, c = x⁴ - 1. Then obviously a > c. Also a - b = (x⁴ - x³) + (x² - x) > 0. But a - b = x⁴ - (x³ - x²) - x < c. So a is the longest side but shorter than b + c. That is sufficient to ensure that there is a triangle with sides a, b, c.

Thanks to Suat Namli

17th VMO 1979

© John Scholes
jscholes@kalva.demon.co.uk
6 March 2004
Last corrected/updated 6 Mar 04