Difference between revisions of "2021 Fall AMC 12B Problems/Problem 13"
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Let <math>c = \frac{2\pi}{11}.</math> What is the value of | Let <math>c = \frac{2\pi}{11}.</math> What is the value of | ||
<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}?</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}?</cmath> | ||
− | <math>\textbf{(A)}\ -1 \qquad\textbf{(B)}\ \frac{\sqrt{ | + | |
+ | <math>\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</math> | \frac{10}{11} \qquad\textbf{(E)}\ 1</math> | ||
− | ==Solution | + | ==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> | ||
+ | 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> | ||
+ | ~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 | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | ==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= | + | {{AMC12 box|year=2021 Fall|ab=B|num-a=14|num-b=12}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:52, 30 July 2024
Problem
Let What is the value of
Solution
Plugging in , we get Since and we get
~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
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 |
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