32nd CanMO 2000 Problem 5 A non-increasing sequence of 100 non-negative reals has the sum of the first two terms at most 100 and the sum of the remaining terms at most 100. What is the largest possible value for the sum of the squares of the terms?
Let the sequence by x 1 , x 2 , ... , x 100 . Given any sequence satisfying the conditions with the sum of the first two terms less than 100, we can obtain another sequence which also satisfies the conditions and has larger sum of squares by increasing the first term. So we may assume that x 1 + x 2 = 100. So the sum of squares is at most (100 - x 2 ) 2 + x 2 2 + ... + x 100 2 = 10000 + 2x 2 2 - 200x 2 + x 3 2 + ... + x 100 2 <= 10000 + 2x 2 2 - x 2 (x 1 + x 2 + ... + x 100 ) + x 3 2 + ... + x 100 2 = 10000 + x 2 (x 2 - x 1 ) + x 3 (x 3 - x 2 ) + ... + x 100 (x 100 - x 2 ). Hence the maximum value is 10000 and is achieved iff all of x 2 (x 2 - x 1 ), x 3 (x 3 - x 2 ), ... , x 100 (x 100 - x 2 ) are zero. x 2 (x 2 - x 1 ) = 0 implies either x 2 = 0 or x 2 = x 1 . The former gives the unique solution x 1 = 100, other terms zero. The latter implies x 1 = x 2 = 50 and every other term must be 0 or 50, which gives the unique solution 50, 50, 50, 50, 0, ... , 0. 32nd CanMO 2000 © John Scholes jscholes@kalva.demon.co.uk 15 June 2002