PUMaC 2011 · 代数（B 组） · 第 1 题
PUMaC 2011 — Algebra (Division B) — Problem 1
题目详情
- [ 3 ] If we define ⊗ ( a, b, c ) by max( a, b, c ) − min( a, b, c ) ⊗ ( a, b, c ) = , a + b + c − min( a, b, c ) − max( a, b, c ) compute ⊗ ( ⊗ (7 , 1 , 3) , ⊗ ( − 3 , − 4 , 2) , 1). 2 8 8
解析
- − 1) − 1 = − 1 = , so | a + b | = 2 · 527 = 1054 . 4 625 5 2 Third Solution : As before, the two roots are 1 ± 2 i . Then, squaring three times, (1 + 2 i ) = 4 8 8 − 3 + 4 i , (1 + 2 i ) = − 7 − 24 i , (1 + 2 i ) = − 527 + 336 i . Similarly, (1 − 2 i ) = − 527 − 336 i (by 8 8 taking the conjugate of both sides), so | a + b | = 2 · 527 = 1054 .