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

(Solution 2)
(Solution 3)
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==Solution 3==
 
==Solution 3==
Using information from Solution 2, we have that <math>5^2 \equiv 3 \pmod{11}</math>, so 3 is a quadratic residue mod 11, and for quadratic residues, their Legendre symbol is <math>\boxed{\textbf{(E)}\ 1}</math>
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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 is the answer from Solution 2 is <math>\boxed{\textbf{(E)}\ 1}</math>
  
 
==Video Solution (Just 2 min!)==
 
==Video Solution (Just 2 min!)==

Revision as of 18:52, 30 July 2024

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 is the answer from Solution 2 is $\boxed{\textbf{(E)}\ 1}$

Video Solution (Just 2 min!)

https://youtu.be/S44IzCpzTeg

~Education, the Study of Everything

See Also

2021 Fall AMC 12B (ProblemsAnswer KeyResources)
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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