Difference between revisions of "2021 Fall AMC 12B Problems/Problem 13"
Lopkiloinm (talk | contribs) (→Solution 2) |
Lopkiloinm (talk | contribs) (→Solution 2) |
||
| Line 17: | Line 17: | ||
==Solution 2== | ==Solution 2== | ||
| − | Let <math>c=\frac{2\pi}{p}</math> where <math>p</math> is an odd prime number. | + | 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)\ldots\sin(\frac{p-1}{2}qc)}{\sin(c)\sin(2c)\ldots\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> | Then <math>\dfrac{\sin(qc)\sin(2qc)\ldots\sin(\frac{p-1}{2}qc)}{\sin(c)\sin(2c)\ldots\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> | ||
Revision as of 22:27, 16 November 2022
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
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
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
