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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 13"

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\frac{10}{11} \qquad\textbf{(E)}\ 1</math>
 
\frac{10}{11} \qquad\textbf{(E)}\ 1</math>
  
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==Solution 1==
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==Solution==
  
 
Plugging in <math>c</math>, we get
 
Plugging in <math>c</math>, we get
 
<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>
 
<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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−
 
 
Since <math>\sin(x+2\pi)=\sin(x),</math> <math>\sin(2\pi-x)=\sin(-x),</math> and <math>\sin(-x)=-\sin(x),</math> we get
 
Since <math>\sin(x+2\pi)=\sin(x),</math> <math>\sin(2\pi-x)=\sin(-x),</math> and <math>\sin(-x)=-\sin(x),</math> we get
 
<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>
 
<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>
  
 
~kingofpineapplz
 
~kingofpineapplz
 +
~Ziyao7294 (minor edit)
 +
 +
==Solution 2==
 +
Eisenstein used such a quotient in his proof of [[quadratic reciprocity]]. Let <math>c=\frac{2\pi}{p}</math> where <math>p</math> is an odd prime number and <math>q</math> is any integer.
 +
 +
Then <math>\dfrac{\sin(qc)\sin(2qc)\cdots\sin(\frac{p-1}{2}qc)}{\sin(c)\sin(2c)\cdots\sin(\frac{p-1}{2}c)}</math> is the Legendre symbol <math>\left(\frac{q}{p}\right)</math>. Legendre symbol is calculated using quadratic reciprocity which is <math>\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}</math>. The Legendre symbol <math>\left(\frac{3}{11}\right)=(-1)\left(\frac{11}{3}\right)=(-1)\left(\frac{-1}{3}\right)=(-1)(-1)=\boxed{\textbf{(E)}\ 1}</math>
 +
 +
~Lopkiloinm
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 +
==Solution 3==
 +
We have that <math>5^2 \equiv 3 \pmod{11}</math>, so 3 is a quadratic residue mod 11. For quadratic residues, their Legendre symbol which we know is the answer from Solution 2 is <math>\boxed{\textbf{(E)}\ 1}</math>
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 +
==Solution 4==
 +
We have <math>\zeta</math> be an 11th primitive root of unity. Then the quotient becomes <cmath>\frac{(\zeta^3-\zeta^{-3})(\zeta^6-\zeta^{-6})(\zeta^9-\zeta^{-9})(\zeta^{12}-\zeta^{-12})(\zeta^{15}-\zeta^{-15})}{(\zeta^1-\zeta^{-1})(\zeta^2-\zeta^{-2})(\zeta^3-\zeta^{-3})(\zeta^{4}-\zeta^{-4})(\zeta^{5}-\zeta^{-5})}</cmath> which we can use modular arithmetic to become <cmath>\frac{(\zeta^3-\zeta^{-3})(\zeta^{-5}-\zeta^{5})(\zeta^{-2}-\zeta^{2})(\zeta^{1}-\zeta^{-1})(\zeta^{4}-\zeta^{-4})}{(\zeta^1-\zeta^{-1})(\zeta^2-\zeta^{-2})(\zeta^3-\zeta^{-3})(\zeta^{4}-\zeta^{-4})(\zeta^{5}-\zeta^{-5})}</cmath> and we see that is <math>\boxed{\textbf{(E)}\ 1}</math>
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 +
~Lopkiloinm
 +
 +
==Video Solution (Just 2 min!)==
 +
https://youtu.be/S44IzCpzTeg
 +
 +
~<i>Education, the Study of Everything</i>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=14|num-b=12}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=14|num-b=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:05, 1 February 2025

Problem

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

Plugging in $c$, we get \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.\] Since $\sin(x+2\pi)=\sin(x),$ $\sin(2\pi-x)=\sin(-x),$ and $\sin(-x)=-\sin(x),$ we get \[\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{\textbf{(E)}\ 1}.\]

~kingofpineapplz ~Ziyao7294 (minor edit)

Solution 2

Eisenstein used such a quotient in his proof of quadratic reciprocity. Let $c=\frac{2\pi}{p}$ where $p$ is an odd prime number and $q$ is any integer.

Then $\dfrac{\sin(qc)\sin(2qc)\cdots\sin(\frac{p-1}{2}qc)}{\sin(c)\sin(2c)\cdots\sin(\frac{p-1}{2}c)}$ is the Legendre symbol $\left(\frac{q}{p}\right)$. Legendre symbol is calculated using quadratic reciprocity which is $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}$. The Legendre symbol $\left(\frac{3}{11}\right)=(-1)\left(\frac{11}{3}\right)=(-1)\left(\frac{-1}{3}\right)=(-1)(-1)=\boxed{\textbf{(E)}\ 1}$

~Lopkiloinm

Solution 3

We have that $5^2 \equiv 3 \pmod{11}$, so 3 is a quadratic residue mod 11. For quadratic residues, their Legendre symbol which we know is the answer from Solution 2 is $\boxed{\textbf{(E)}\ 1}$

Solution 4

We have $\zeta$ be an 11th primitive root of unity. Then the quotient becomes \[\frac{(\zeta^3-\zeta^{-3})(\zeta^6-\zeta^{-6})(\zeta^9-\zeta^{-9})(\zeta^{12}-\zeta^{-12})(\zeta^{15}-\zeta^{-15})}{(\zeta^1-\zeta^{-1})(\zeta^2-\zeta^{-2})(\zeta^3-\zeta^{-3})(\zeta^{4}-\zeta^{-4})(\zeta^{5}-\zeta^{-5})}\] which we can use modular arithmetic to become \[\frac{(\zeta^3-\zeta^{-3})(\zeta^{-5}-\zeta^{5})(\zeta^{-2}-\zeta^{2})(\zeta^{1}-\zeta^{-1})(\zeta^{4}-\zeta^{-4})}{(\zeta^1-\zeta^{-1})(\zeta^2-\zeta^{-2})(\zeta^3-\zeta^{-3})(\zeta^{4}-\zeta^{-4})(\zeta^{5}-\zeta^{-5})}\] and we see that is $\boxed{\textbf{(E)}\ 1}$

~Lopkiloinm

Video Solution (Just 2 min!)

https://youtu.be/S44IzCpzTeg

~Education, the Study of Everything

See Also

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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