7th USAMO 1978 Problem 1 The sum of 5 real numbers is 8 and the sum of their squares is 16. What is the largest possible value for one of the numbers? Solution
: 16/5. Let the numbers be v, w, x, y, z. We have (v -6/5) 2 + (w - 6/5) 2 + ... + (z - 6/5) 2 = (v 2 + ... + z 2 ) - 12/5 (v + ... + z) + 36/5 = 16 - 96/5 + 36/5 = 4. Hence |v - 6/5| ≤ 2, so v ≤ 16/5. This value can be realized by putting v = 16/5 and setting the other numbers to 6/5. 7th USAMO 1978 © John Scholes jscholes@kalva.demon.co.uk 11 May 2002