HMMT 二月 2013 · 代数 · 第 8 题

英文解析

8. Let x, y be complex numbers such that = 4 and = 2. Find all possible values of .
   3 3 5 5
   x + y x + y x + y
   √
   1 1 xy
   Answer: 10 ± 2 17 Let A = + and let B = + . Thenx y y x
   2 2
   B x + y = = 4 ,
   A x + yso B = 4 A . Next, note that
   4 4 3 3
   x + y x + y
   B − 2 = and AB − A = ,2
   2 2 2 2
   x y x yso
   B − 22 = 2 .
   AB − A
   √
   − 1 ± 17
   Substituting B = 4 A and simplifying, we find that 4 A + A − 1 = 0, so A = . Finally, note that 2
   6 6 5 58
   x + y x + y
   3 3 3 2 2
   64 A − 12 A = B − 3 B = and 16 A − 4 A − A = A ( B − 2) − ( AB − A ) = ,
   3 3 3 3
   x y x yso
   Algebra Test
   6 6 2
   x + y 64 A − 12 4 − 16 A = = ,
   5 5 2
   x + y 16 A − 4 A − 1 3 − 8 A
   √
   − 1+ 17
   where the last inequality follows from the fact that 4 A = 1 − A . If A = , then this value equals 2
   √8
   √ √
   − 1 − 17
   10 + 2 17. Similarly, if A = , then this value equals 10 − 2 17.
   (It is not hard to see that these values are achievable by noting that with the values of A and B we 8
   can solve for x + y and xy , and thus for x and y .)
   23