2021 AMC 12A Problems/Problem 22

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Problem

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

$\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}$

Solution

Part 1: solving for $\cos \frac{6\pi}7 = \cos \frac{8\pi}7$

$c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$

Multiply by $8 \sin{2\pi}7$

$c 8 \sin{2\pi}7 = -8 \sin{2\pi}7 \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$

Then use sine addition formula backwards:

$2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7$

$c 8 \sin{2\pi}7 = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$

$c 8 \sin{2\pi}7 = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7$

$c 8 \sin{2\pi}7 = -\sin \frac{16\pi}7$

$c 8 \sin{2\pi}7 = -\sin \frac{2\pi}7$

$c = -\frac{1}8$


This is in progress - I am working on solutions for a and b.

~Tucker

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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