HMMT 二月 2011 · TEAM2 赛 · 第 3 题

HMMT February 2011 — TEAM2 Round — Problem 3

[ 30 ] Let x and y be complex numbers such that | x | = | y | = 1. (a) [ 15 ] Determine the maximum value of | 1 + x | + | 1 + y | − | 1 + xy | . 2 2011 (b) [ 15 ] Determine the maximum value of | 1 + x | + | 1 + xy | + | 1 + xy | + . . . + | 1 + xy | − 1006 | 1 + y | .

解析

[ 30 ] Let x and y be complex numbers such that | x | = | y | = 1. (a) [ 15 ] Determine the maximum value of | 1 + x | + | 1 + y | − | 1 + xy | . 2 2011 (b) [ 15 ] Determine the maximum value of | 1 + x | + | 1 + xy | + | 1 + xy | + . . . + | 1 + xy | − 1006 | 1 + y | . Solution: √ (a) Answer: 2 2 √ (b) Answer: 2012 2 We divide the terms into 1006 sums of the form 2 k 2 k +1 | 1 + xy | + | 1 + xy | − | 1 + y | . Team Round B For each of these, we obtain, as in part a, ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 2 k 2 k +1 2 k 2 k +1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 + xy + 1 + xy − | 1 + y | ≤ 1 + xy + xy − y ∣ ∣ ∣ ∣ 2 k 2 k ∣ ∣ ∣ ∣ = 1 + xy + | y | 1 − xy ∣ ∣ ∣ ∣ 2 k 2 k ∣ ∣ ∣ ∣ = 1 + xy + 1 − xy √ 2 4 k ≤ 2 | 1 − x y | √ 2 4 Again, this is maximized when x y = − 1, which leaves a total sum of at most 2012 2. If we let √ √ x = i and y = − 1, every sum of three terms will be 2 2, for a total of 2012 2 , as desired.