12th Chinese 1997 Problem A1 The real numbers x 1 , x 2 , ... , x 1997 have sum -318 √3 and satisfy -1/√3 ≤ x i ≤ √3. What is the maximum possible value for the sum of their 12th powers? Solution
: 1736/3 6 + 260.3 6 + 2 12 /3 6 . (x + y) 12 + (x - y) 12 is an increasing function of y, so we maximise the sum by taking as many as possible of the numbers at the endpoints of the allowed interval. There certainly cannot be more than two numbers not at the endpoints, for otherwise we could increase one and decrease the other, thus increasing the sum. If we take n equal to -1/√3 and 1997 - n equal to √3, then -n/√3 + (1997 - n)√3 = -318√3, so n = 1736.25, which does not work. So we cannot have all the numbers at the endpoints. But we can take 1736 equal to -1/√3, 260 equal to √3 and 1 equal to 2/√3. That gives 1736/3 6 + 260.3 6 + 2 12 /3 6 for the sum. But we need to check that this is the only possibility for just one number not at the endpoints. Suppose we take n at -1/√3 and 1996 - n at √3. Then the other number must be -318√3 + n/√3 + (n - 1996)√3 = -2413√3 + 4/3 n √3. This must lie between -1/√3 and √3, so 4n/3 - 2314 must lie between -1/3 and 1. Hence n lies between 1735.25 and 1736.25, so the only possibility is 1736. 12th Chinese 1997 © John Scholes jscholes@kalva.demon.co.uk 5 June 2002