HMMT 二月 2015 · 冲刺赛 · 第 26 题

英文解析

26. [ 17 ] Let a = 17 and b = i 19, where i = − 1. Find the maximum possible value of the ratio
    | a − z | / | b − z | over all complex numbers z of magnitude 1 (i.e. over the unit circle | z | = 1).
    Answer: Let | a − z | / | b − z | = k. We wish to determine the minimum and maximum value of k .4
    Squaring and expansion give:3
    2 2 2
    | a − z | = | b − z | · k
    2 2 2
    | a | − 2 a · z + 1 = ( | b | − 2 b · z + 1) k
    2 2 2 2
    | a | + 1 − ( | b | + 1) k = 2( a − bk ) · z,
    where · is a dot product of complex numbers, i.e., the dot product of vectors corresponding to thecomplex numbers in the complex plane. Now, since z has modulus 1 but can assume any direction,
    the only constraint on the value of k is
    2 2 2 2
    || a | + 1 − ( | b | + 1) k | ≤ 2 | a − bk | .
    Squaring again and completing the square, the inequality reduces to:
    2 2 2 2 4 2 2 2
    ( | a | − 1) + ( | b | − 1) k + 2(4 a · b − ( | a | + 1)( | b | + 1)) k ≤ 0
    2 2 2 2 2 2
    (( | a | − 1) − ( | b | − 1) k ) − 4 | a − b | k ≤ 0
    2 2 2
    | ( | a | − 1) − ( | b | − 1) k | ≤ 2 | a − b | k.
    At this stage all the relevant expressions are constant real numbers. Denote, for simplicity, A =
    2 2 2
    | a | − 1 , B = | b | − 1, and C = | a − b | . Then, we are looking for k such that | A − Bk | ≤ 2 Ck . If B = 0,
    A Athen k ≥ | | , so the minimum value is | | and the maximum value is + ∞ . Otherwise, consider
    2 C 2 C
    2 2 2 2 2
    C + AB = ( | a | − 2 a · b + | b | ) + ( | a | − 1)( | b | − 1) = | ab | − 2 a · b + 12 = | ab | − 2 ℜ ( ab ) + 12 = | ab − 1 |2
    √
    So let D = | ab − 1 | = C + AB . We may assume B > 0 (the another case is analogous: just substitute 2
    A, B with − A, − B ). Then, k is determined by the following inequalities:
    Bk + 2 Ck − A ≥ 02
    Bk − 2 Ck − A ≤ 02 − C − D − C + D C − D C + D
    The first inequality gives k ≤ or k ≥ , and the second gives ≤ k ≤ . Combining,
    B B B B
    C − D C + Dthis gives | | ≤ k ≤ | | , as claimed.
    B B
    2 2
    To summarize the general answer, let A = | a | − 1 , B = | b | − 1 , C = | a − b | , D = | ab − 1 | . Then, if
    A C − D C + D
    | b | = 1, min is | | and max is + ∞ ; otherwise, min is | | and max is | | .
    2 C B B
    √ √ ∣ √ √ ∣ √
    ∣ ∣
    In the special case a = 17 and b = 19 i , we have A = 16, B = 18, C = 17 − 19 i = 36 = 6,
    √
    C + D 6+18 4
    and D = 17 · 19 + 1 = 18. Thus the answer is = = .
    B 18 3