Question

The polynomial $\begin{aligned} P(x) &= \small (1 + x + x^2 + \cdots + x^{17})^2 \\ &\quad {}- x^{17} \end{aligned}$ has $34$ complex zeros of the form $z_k$ $= r_k[\cos(2\pi\alpha_k) + i\sin(2\pi\alpha_k)],$ $k = 1, 2, 3, \ldots, 34,$ with $0 \lt \alpha_1 \le \alpha_2 \le \alpha_3$ $\le \cdots \le \alpha_{34} \lt 1$ and $r_k \gt 0.$ Given that $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 + \alpha_5 = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Answer

Answer: 482

Concepts:roots of unity geometric sequence factoring

Difficulty rating: 3060

Solution:

For $x \ne 1,$ write $1 + x + \cdots + x^{17} = \frac{x^{18} - 1}{x - 1},$ so $\begin{aligned} \scriptsize (x - 1)^2 P(x) &= (x^{18} - 1)^2 \\ &\quad {}- x^{17}(x - 1)^2 \\ &= x^{36} - x^{19} - x^{17} + 1 \\ &= (x^{19} - 1)(x^{17} - 1). \end{aligned}$ Hence the zeros of $P$ are the $34$ complex numbers other than $1$ satisfying $x^{17} = 1$ or $x^{19} = 1;$ all lie on the unit circle, with angles $\alpha = \frac{k}{17}$ $(k = 1, \ldots, 16)$ and $\alpha = \frac{k}{19}$ $(k = 1, \ldots, 18).$

The five smallest of these angles are $\frac{1}{19} \lt \frac{1}{17} \lt \frac{2}{19} \lt \frac{2}{17} \lt \frac{3}{19},$ whose sum is $\frac{6}{19} + \frac{3}{17} = \frac{102 + 57}{323} = \frac{159}{323}.$ Since $159 = 3 \cdot 53$ and $323 = 17 \cdot 19,$ this is in lowest terms, and $m + n = 159 + 323 = 482.$

2004 AIME I Problem 13

Answer: 482

Difficulty rating: 3060

Solution:

Problem 13 in Other Years