2021 AMC 12A Problem 22

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.