Suppose that the roots of the polynomial P(x) = x^3 + ax^2 + bx + c are \cos(2π)/(7), \cos(4π)/(7), and \cos(6π)/(7), where angles are in radians. What is abc?

2021 AMC 12A Spring Problem 22 — Solution

2021 AMC 12A Spring Problem 22

Below is the professionally curated solution for Problem 22 of the 2021 AMC 12A Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Spring solutions, or check the answer key.

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Concepts:roots of unity polynomial Vieta’s Formulas

Difficulty rating: 2450

22.

Suppose that the roots of the polynomial $P(x) = x^3 + ax^2 + bx + c$ are $\cos\tfrac{2\pi}{7},$ $\cos\tfrac{4\pi}{7},$ and $\cos\tfrac{6\pi}{7},$ where angles are in radians. What is $abc?$

$-\dfrac{3}{49}$

$-\dfrac{1}{28}$

$\dfrac{\sqrt[3]{7}}{64}$

$\dfrac{1}{32}$

$\dfrac{1}{28}$

Solution:

The numbers $\cos\tfrac{2\pi}{7},$ $\cos\tfrac{4\pi}{7},$ $\cos\tfrac{6\pi}{7}$ are the three roots of $8x^3 + 4x^2 - 4x - 1 = 0.$ Dividing by $8$ puts it in monic form: $x^3 + \tfrac12 x^2 - \tfrac12 x - \tfrac18 = 0.$

Matching coefficients, $a = \tfrac12,$ $b = -\tfrac12,$ $c = -\tfrac18.$ Therefore $abc = \tfrac12\cdot\left(-\tfrac12\right)\cdot\left(-\tfrac18\right) = \tfrac{1}{32}.$

Thus, the correct answer is D.

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