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HMMT 二月 2009 · 代数 · 第 9 题

HMMT February 2009 — Algebra — Problem 9

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 7 ] Let f ( x ) = x + 14 x + 52 x + 56 x + 16. Let z , z , z , z be the four roots of f . Find the smallest
    1 2 3 4
    possible value of | z z + z z | where { a, b, c, d } = { 1 , 2 , 3 , 4 } .
    a b c d 3
解析

英文解析

  1. [ 7 ] Let f ( x ) = x + 14 x + 52 x + 56 x + 16. Let z , z , z , z be the four roots of f . Find the smallest
    1 2 3 4
    possible value of | z z + z z | where { a, b, c, d } = { 1 , 2 , 3 , 4 } .
    a b c d
    Answer: 8
    1 4 3 2
    Solution: Note that f (2 x ) = x + 7 x + 13 x + 7 x + 1. Because the coefficients of this polynomialare symmetric, if r is a root of f ( x ) then is as well. Further, f ( − 1) = − 1 and f ( − 2) = 16 so f ( x )416
    has two distinct roots on ( − 2 , 0) and two more roots on ( −∞ , − 2). Now, if σ is a permutation ofr
    { 1 , 2 , 3 , 4 } :
    | z z + z z | ≤ ( z z + z z + z z + z z )1
    σ (1) σ (2) σ (3) σ (4) σ (1) σ (2) σ (3) σ (4) σ (4) σ (3) σ (2) σ (1)
    Let the roots be ordered z ≤ z ≤ z ≤ z , then by rearrangement the last expression is at least:2
    1 2 3 4
    ( z z + z z + z z + z z )1
    1 4 2 3 3 2 4 1
    Since the roots come in pairs z z = z z = 4, our expression is minimized when σ (1) = 1 , σ (2) =2
    1 4 2 3
    4 , σ (3) = 3 , σ (4) = 2 and its minimum value is 8. 3