HMMT 二月 2011 · TEAM2 赛 · 第 3 题

英文解析

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 | .
   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.