HMMT 二月 2022 · 冲刺赛 · 第 13 题

HMMT February 2022 — Guts Round — Problem 13

[9] Let z , z , z , z be the solutions to the equation x +3 x +3 x +3 x +1 = 0. Then | z | + | z | + | z | + | z | 1 2 3 4 1 2 3 4 √ a + b c can be written as , where c is a squarefree positive integer, and a, b, d are positive integers with d gcd( a, b, d ) = 1. Compute 1000 a + 100 b + 10 c + d .

解析

[9] Let z , z , z , z be the solutions to the equation x +3 x +3 x +3 x +1 = 0. Then | z | + | z | + | z | + | z | 1 2 3 4 1 2 3 4 √ a + b c can be written as , where c is a square-free positive integer, and a, b, d are positive integers with d gcd( a, b, d ) = 1. Compute 1000 a + 100 b + 10 c + d . Proposed by: Daniel Zhu Answer: 7152 2 Solution: Note that x = 0 is clearly not a solution, so we can divide the equation by x to get √ 2 1 1 1 2 1 − 3 ± 5 x + 2 + +3 x + +1 = 0. Letting y = x + , we get that y +3 y +1 = 0, so y = x + = . 2 x x x x 2 √ − 3+ 5 Since has absolute value less than 2, the associated x are on the unit circle, and thus the two 2 √ − 3 − 5 solutions for x in this case each have magnitude 1. For , the roots are negative reals that are 2 reciprocals of each other. Thus, the sum of their absolute values is the absolute value of their sum, √ √ √ 3+ 5 3+ 5 7+ 5 which is . Thus, the sum of the magnitudes of the four solutions are 1 + 1 + = . 2 2 2