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Difference between revisions of "2021 AMC 12A Problems/Problem 22"

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Part 1: solving for <math>\cos \frac{6\pi}7 = \cos \frac{8\pi}7</math>
 
Part 1: solving for <math>\cos \frac{6\pi}7 = \cos \frac{8\pi}7</math>
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<math>c</math> is the negation of the product of roots by Vieta's formulas
  
 
<math>c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math>
 
<math>c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math>
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−
This is in progress - I am working on solutions for a and b.
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Part 2: in progress - solving for b
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<math>b</math> is the sum of roots two at a time by Vieta's
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<math>b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7</math>
 +
 
 +
 
 +
Part 3: solving for a
 +
 
 +
<math>a</math> is the negation of the sum of roots
 +
 
 +
<math>a = \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7</math>
 +
 
 +
The real values of the 7th roots of unity are: <math>1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7</math> and they sum to <math>0</math>.
 +
 
 +
If we subtract 1, and condense identical terms, we get:
 +
 
 +
<math>2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1</math>
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Therefore, we have $a = -\frac{1}2
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This is in progress - I am working on the solution for b.
  
 
~Tucker
 
~Tucker

Revision as of 16:36, 11 February 2021

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 $\cos \frac{6\pi}7 = \cos \frac{8\pi}7$

$c$ is the negation of the product of roots by Vieta's formulas

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

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:

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

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

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

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

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

$c = -\frac{1}8$


Part 2: in progress - solving for b

$b$ is the sum of roots two at a time by Vieta's

$b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7$


Part 3: solving for a

$a$ is the negation of the sum of roots

$a = \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7$

The real values of the 7th roots of unity are: $1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7$ and they sum to $0$.

If we subtract 1, and condense identical terms, we get:

$2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1$

Therefore, we have $a = -\frac{1}2


This is in progress - I am working on the solution for b.

~Tucker

See also

2021 AMC 12A (Problems • Answer Key • Resources)
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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