HMMT 二月 2021 · ALGNT 赛 · 第 7 题

HMMT February 2021 — ALGNT Round — Problem 7

Suppose that x , y , and z are complex numbers of equal magnitude that satisfy √ √ 3 x + y + z = − − i 5 2 and √ √ xyz = 3 + i 5 . If x = x + ix , y = y + iy , and z = z + iz for real x , x , y , y , z , and z , then 1 2 1 2 1 2 1 2 1 2 1 2 2 ( x x + y y + z z ) 1 2 1 2 1 2 a can be written as for relatively prime positive integers a and b . Compute 100 a + b . b lcm( a,b )

解析

Suppose that x , y , and z are complex numbers of equal magnitude that satisfy √ √ 3 x + y + z = − − i 5 2 and √ √ xyz = 3 + i 5 . If x = x + ix , y = y + iy , and z = z + iz for real x , x , y , y , z , and z , then 1 2 1 2 1 2 1 2 1 2 1 2 2 ( x x + y y + z z ) 1 2 1 2 1 2 a can be written as for relatively prime positive integers a and b . Compute 100 a + b . b Proposed by: Akash Das Answer: 1516 √ Solution: From the conditions, it is clear that a, b, c all have magnitude 2. Conjugating the first √ √ √ √ √ √ ab + bc + ca 3 3 5 equation gives 2( ) = − + i 5, which means ab + bc + ca = ( − + i )( 3 + i 5) = abc 2 4 2 √ − 13+ i 15 . Then, 4 1 2 2 2 a a + b b + c c = Im( a + b + c ) 1 2 1 2 1 2 2 1 2 = Im(( a + b + c ) ) − Im( ab + bc + ca ) 2 √ 15 = , 4 so the answer is 1516. Remark: { } √ √ √ √ √ √ − 3 − i 5 − 3 3 − i 5 3 3 − i 5 { a, b, c } = , , 2 4 4 lcm( a,b )