HMMT 二月 2008 · 冲刺赛 · 第 33 题

英文解析

33. [ 18 ] Let a , b , c be nonzero real numbers such that a + b + c = 0 and a + b + c = a + b + c . Find
    2 2 2
    the value of a + b + c .
    Answer: Let σ = a + b + c , σ = ab + bc + ca and σ = abc be the three elementary symmetric 6
    1 2 3
    3 3 35
    polynomials. Since a + b + c is a symmetric polynomial, it can be written as a polynomial in σ , σ
    1 2
    and σ . Now, observe that σ = 0, and so we only need to worry about the terms not containing σ . By
    3 1 1
    3 3 3
    considering the degrees of the terms, we see that the only possibility is σ . That is, a + b + c = kσ
    3 3
    for some constant k . By setting a = b = 1, c = − 2, we see that k = 3.
    5 5 5
    By similar reasoning, we find that a + b + c = hσ σ for some constant h . By setting a = b = 1 and
    2 3
    c = − 2, we get h = − 5.
    So, we now know that a + b + c = 0 implies 9
    3 3 3 5 5 5
    a + b + c = 3 abc and a + b + c = − 5 abc ( ab + bc + ca )
    3 3 3 5 5 5
    Then a + b + c = a + b + c implies that 3 abc = − 5 abc ( ab + bc + ca ). Given that a, b, c are nonzero,
    3 6
    2 2 2 2
    we get ab + bc + ca = − . Then, a + b + c = ( a + b + c ) − 2( ab + bc + ca ) = .
    5 5
    11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUNDth