2021 AMC 12A Spring Problem 22

Question

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}

The three cosines are the roots of

8x^3 + 4x^2 - 4x - 1 = 0

Divide by

8

to match

x^3 + ax^2 + bx + c,

reading off

a, b, c

directly

Let the three roots be

r_1,r_2,r_3.

The seventh roots of unity give

r_1+r_2+r_3=-\tfrac12.

Doubling the three angles merely permutes their cosines, so

\sum r_i^2=\tfrac32+\tfrac12\sum r_i=\tfrac54.

Hence

r_1r_2+r_1r_3+r_2r_3

equals

\tfrac12((\sum r_i)^2-\sum r_i^2)=-\tfrac12.

The product-to-sum identity gives

4r_1r_2r_3=1+r_1+r_2+r_3=\tfrac12,

so

r_1r_2r_3=\tfrac18.

Therefore their monic polynomial is

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.

Problem 22 in other years