HMMT 二月 2013 · 代数 · 第 8 题

HMMT February 2013 — Algebra — Problem 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 23

解析

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