Difference between revisions of "2021 Fall AMC 12B Problems/Problem 13"
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==Solution 2== | ==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)\ | + | 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> | ||
| + | |||
| + | ==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> | ||
~Lopkiloinm | ~Lopkiloinm | ||
| − | |||
==Video Solution (Just 2 min!)== | ==Video Solution (Just 2 min!)== | ||
Latest revision as of 03:05, 1 February 2025
Contents
Problem
Let 
What is the value of
![\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]](http://latex.artofproblemsolving.com/5/c/b/5cb66360bc69b72e86eac3564c26bae7fd577bc9.png)
Solution
Plugging in 
, we get
![\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.\]](http://latex.artofproblemsolving.com/f/f/8/ff84ea1918eb8039edc31877eb63ca11dfef1986.png)
Since 
and 
we get
![\[\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{\textbf{(E)}\ 1}.\]](http://latex.artofproblemsolving.com/e/7/e/e7e5d4583710682a595b107182f40d55b4cad8b7.png)
~kingofpineapplz ~Ziyao7294 (minor edit)
Solution 2
Eisenstein used such a quotient in his proof of quadratic reciprocity. Let 
where 
is an odd prime number and 
is any integer.
Then 
is the Legendre symbol 
. Legendre symbol is calculated using quadratic reciprocity which is 
. The Legendre symbol 
~Lopkiloinm
Solution 3
We have that 
, so 3 is a quadratic residue mod 11. For quadratic residues, their Legendre symbol which we know is the answer from Solution 2 is 
Solution 4
We have 
be an 11th primitive root of unity. Then the quotient becomes ![\[\frac{(\zeta^3-\zeta^{-3})(\zeta^6-\zeta^{-6})(\zeta^9-\zeta^{-9})(\zeta^{12}-\zeta^{-12})(\zeta^{15}-\zeta^{-15})}{(\zeta^1-\zeta^{-1})(\zeta^2-\zeta^{-2})(\zeta^3-\zeta^{-3})(\zeta^{4}-\zeta^{-4})(\zeta^{5}-\zeta^{-5})}\]](http://latex.artofproblemsolving.com/2/3/8/238aa3463016f1406f33d68e25ff8f077d5ca03a.png)
which we can use modular arithmetic to become ![\[\frac{(\zeta^3-\zeta^{-3})(\zeta^{-5}-\zeta^{5})(\zeta^{-2}-\zeta^{2})(\zeta^{1}-\zeta^{-1})(\zeta^{4}-\zeta^{-4})}{(\zeta^1-\zeta^{-1})(\zeta^2-\zeta^{-2})(\zeta^3-\zeta^{-3})(\zeta^{4}-\zeta^{-4})(\zeta^{5}-\zeta^{-5})}\]](http://latex.artofproblemsolving.com/a/2/e/a2e6d07343b0ffc8642bd7ae054a801b3b8ec182.png)
and we see that is
~Lopkiloinm
Video Solution (Just 2 min!)
~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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
