HMMT 二月 2014 · 冲刺赛 · 第 32 题
HMMT February 2014 — Guts Round — Problem 32
题目详情
- [ 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND Organization Team Team ID# 2 2
解析
- [ 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 i which is equivalent to a + bi = by the quadratic formula. 2 10 i Similarly, the second equation plus i times the first yields 4 + 5 i = b + ai − , which is equivalent b + ai 4+(5 ± 3) i to b + ai = . 2 1 1 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 √ √ 2 ( a − 1)( a − 4) for some ∈ {− 1 , 1 } , and substituting in and expanding gives 0 = ( − 2 a +5 a ) ( a − 1)( a − 4). λ More symmetrically, we may write a = λ (4 − b ), b = λ (5 − a ) to get ( a, b ) = (4 − 5 λ, 5 − 4 λ ), and 2 1 − λ 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