HMMT 二月 2013 · 冲刺赛 · 第 2 题

英文解析

2. [ 4 ] The real numbers x, y, z satisfy 0 ≤ x ≤ y ≤ z ≤ 4. If their squares form an arithmetic progressionwith common difference 2, determine the minimum possible value of | x − y | + | y − z | .
   √
   2 2
   z − x 4
   Answer: 4 − 2 3 Clearly | x − y | + | y − z | = z − x = = , which is minimized when z = 4
   z + x z + x
   √ √ √
   and x = 12. Thus, our answer is 4 − 12 = 4 − 2 3.