1/√2 Solution It is easy to check that x 1 = x 5 = 0, x 2 = x 4 = 1, x 3 = √2, t = 1/√2 works. Suppose for some t we have a solution x i . Then 0 = 2tx 2 + 2tx 4 - 4t 2 x 3 = x 1 + 2x 3 + x 5 - 4t 2 x 3 . Hence x 3 = 0 or 4t 2 - 2 = (x 1 + x 5 )/x 3 ≥ 0. If x 3 = 0, then x 2 + x 4 = 0, so x 2 = x 4 = 0. Hence x 1 + x 3 = 0 and x 3 + x 5 = 0, so all x i = 0. Contradiction. So we must have 4t 2 - 2 ≥ 0 and hence t ≥ 1/√2. 14th Swedish 1974 © John Scholes jscholes@kalva.demon.co.uk 13 February 2004 Last corrected/updated 13 Feb 04
14th Swedish 1974 Problem 5 Find the smallest positive real t such that x 1 + x 3 = 2t x 2 , x 2 + x 4 = 2t x 3 , x 3 + x 5 = 2t x 4 has a solution x 1 , x 2 , x 3 , x 4 , x 5 in non-negative reals, not all zero.