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

(Added solution for c, feel free to add on, I will be working on a and b right now)
m (Latex fixes. I am not sure if my browser is being weird or some eqns arent rendering)
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Then use sine addition formula backwards:
 
Then use sine addition formula backwards:
  
−
<math>2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7
+
<math>2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7</math>
  
−
</math>c 8 \sin{2\pi}7 = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7<math>
+
<math>c 8 \sin{2\pi}7 = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math>
  
−
</math>c 8 \sin{2\pi}7 = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7<math>
+
<math>c 8 \sin{2\pi}7 = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7</math>
  
−
</math>c 8 \sin{2\pi}7 = -\sin \frac{16\pi}7<math>
+
<math>c 8 \sin{2\pi}7 = -\sin \frac{16\pi}7</math>
  
−
</math>c 8 \sin{2\pi}7 = -\sin \frac{2\pi}7<math>
+
<math>c 8 \sin{2\pi}7 = -\sin \frac{2\pi}7</math>
  
−
</math>c = -\frac{1}8$
+
<math>c = -\frac{1}8</math>
  
  
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~Tucker
 
~Tucker
−
 
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}}
 
{{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:27, 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 $\cos \frac{6\pi}7 = \cos \frac{8\pi}7$

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

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 (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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