HMMT 二月 2014 · 冲刺赛 · 第 32 题

英文解析

32. [ 20 ] Find all ordered pairs ( a, b ) of complex numbers with a + b 6 = 0, a + = 5, and b + = 4.
    2 2 2 2
    a + b a + b
    5 3
    Answer: (1 , 2) , (4 , 2) , ( , 2 ± i ) Solution 1. First, it is easy to see that ab 6 = 0. Thus, we can
    2 2
    write
    5 − a 4 − b 10 = = .
    2 2
    b a a + b
    Then, we have
    10 4 a − ab 5 b − ab 4 a + 5 b − 2 ab = = = .
    2 2 2 2 2 2
    a + b a b a + b
    Therefore, 4 a + 5 b − 2 ab = 10, so (2 a − 5)( b − 2) = 0. Now we just plug back in and get the four
    5 3
    solutions: (1 , 2) , (4 , 2) , ( , 2 ± i ). It’s not hard to check that they all work.
    2 2
    10( b + ai )
    10 i
    Solution 2. The first equation plus i times the second yields 5 + 4 i = a + bi + = a + bi − ,
    2 2
    a + b a + bi
    (5 ± 3)+4 iwhich is equivalent to a + bi = by the quadratic formula.
    10 i 2
    Similarly, the second equation plus i times the first yields 4 + 5 i = b + ai − , which is equivalentb + ai
    4+(5 ± 3) ito b + ai = .
    1 12
    Letting , ∈ {− 1 , 1 } be the signs in a + bi and b + ai , we get ( a, b ) = ( a + bi, b + ai ) − i ( b + ai, a + bi ) =
    1 2
    2 2
    10+( + )3 8+( − )3 i
    1 2 2 1
    ( , ).
    4 4
    5 − a 4 − b
    Comment. Many alternative approaches are possible. For instance, = = ⇒ b − 2 =
    b a
    √ √
    ( a − 1)( a − 4) for some ∈ {− 1 , 1 } , and substituting in and expanding gives 0 = ( − 2 a +5 a ) ( a − 1)( a − 4).2
    More symmetrically, we may write a = λ (4 − b ), b = λ (5 − a ) to get ( a, b ) = (4 − 5 λ, 5 − 4 λ ), andλ
    1 − λ2
    2 2 4 3 2 2
    then plug into a + b = 10 λ to get 0 = 10( λ + 1) − 41( λ + λ ) + 60 λ = ( λ − 2)(2 λ − 1)(5 λ − 8 λ + 5).
    2 2