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

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~kingofpineapplz
~kingofpineapplz
~Ziyao7294
~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
 
==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>
 
==Solution 4==
We have <imath>\zeta</imath> 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 <imath>\boxed{\textbf{(E)}\ 1}</imath>
 
~Lopkiloinm
 
==Solution 5 (guess)==
 
Since we can't calculate the exact value of <imath>\sin{\frac{2\pi}{11}},</imath> it's unlikely the answer will have radicals in it. This leaves us with <imath>-1, 1,</imath> and <imath>\frac{10}{11}.</imath> The fraction is too precise for it to be a likely answer, so that leaves us with <imath>1</imath> or <imath>-1.</imath> Note that if the angle is greater than <imath>\pi,</imath> the sine will be negative, and if it's less than <imath>\pi</imath> it'll be positive. The numerator and denominator are therefore both positive, so we eliminate <imath>-1</imath> to get <imath>\boxed{\textbf{(E)}\ 1}.</imath>
 
~[[User:grogg007|grogg007]]
 
==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 14:17, 9 November 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

Solution 5 (guess)

Since we can't calculate the exact value of $\sin{\frac{2\pi}{11}},$ it's unlikely the answer will have radicals in it. This leaves us with $-1, 1,$ and $\frac{10}{11}.$ The fraction is too precise for it to be a likely answer, so that leaves us with $1$ or $-1.$ Note that if the angle is greater than $\pi,$ the sine will be negative, and if it's less than $\pi$ it'll be positive. The numerator and denominator are therefore both positive, so we eliminate $-1$ to get $\boxed{\textbf{(E)}\ 1}.$

~grogg007

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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