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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 12"

(Created page with "==Problem 12== Let <math>c = \frac{2\pi}{11}.</math> What is the value of <cmath>\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c...")
 
(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
  
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Plugging in <math>c</math>, we get
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The prime factorization of <math>768</math> is <math>2^8*3</math> and the prime factorization of <math>384</math> is <math>2^7*3</math> so
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<cmath>\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.</cmath>
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<cmath>f(768)=(1+\frac{1}{2}+\ldots+\frac{1}{256})(1+\frac{1}{3})=\frac{511}{192}</cmath>
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<cmath>f(768)=(1+\frac{1}{2}+\ldots+\frac{1}{128})(1+\frac{1}{3})=\frac{510}{192}</cmath>
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so the difference is <math>\boxed{(B) \frac{1}{192}}</math>
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Since <math>\sin(x+2\pi)=\sin(x),</math> <math>\sin(2\pi-x)=\sin(-x),</math> and <math>\sin(-x)=-x,</math> we get
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~lopkiloinm
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<cmath>\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{\textbf{(E)}\ 1}.</cmath>
 
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~kingofpineapplz
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:51, 24 November 2021

Problem 12

Let $c = \frac{2\pi}{11}.$ What is the value of \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\] $\textbf{(A)}\ -1 \qquad\textbf{(B)}\ \frac{\sqrt{-11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{10}{11} \qquad\textbf{(E)}\ 1$

Solution 1

The prime factorization of $768$ is $2^8*3$ and the prime factorization of $384$ is $2^7*3$ so \[f(768)=(1+\frac{1}{2}+\ldots+\frac{1}{256})(1+\frac{1}{3})=\frac{511}{192}\] \[f(768)=(1+\frac{1}{2}+\ldots+\frac{1}{128})(1+\frac{1}{3})=\frac{510}{192}\] so the difference is $\boxed{(B) \frac{1}{192}}$ ~lopkiloinm

See Also

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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