Difference between revisions of "2020 AMC 12A Problems/Problem 22"
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Aha! <math>\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} </math> is a geometric sequence that evaluates to <math>\frac{1}{1-\frac{5}{7}e^{2\theta i}}</math>! Now we can quickly see that <cmath>\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5},</cmath> <cmath>\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}.</cmath> Therefore, <cmath>\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i.</cmath> The imaginary part is <math>\frac{7}{8}</math>, so our answer is <math>\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}} \Rightarrow \textbf{(B)}</math>. | Aha! <math>\sum_{n=0}^\infty \left( \frac{5}{7} \right)^ne^{2i\theta n} </math> is a geometric sequence that evaluates to <math>\frac{1}{1-\frac{5}{7}e^{2\theta i}}</math>! Now we can quickly see that <cmath>\sin(2\theta) = 2 \cdot \sin \theta \cdot \cos \theta = 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \frac{4}{5},</cmath> <cmath>\cos (2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{4}{5}-\frac{1}{5} = \frac{3}{5}.</cmath> Therefore, <cmath>\frac{1}{1-\frac{5}{7}e^{2\theta i}} = \frac{1}{1 - \frac{5}{7}\left( \frac{3}{5} + \frac{4}{5}i\right)} = \frac{7}{8} + \frac{7}{8}i.</cmath> The imaginary part is <math>\frac{7}{8}</math>, so our answer is <math>\frac{1}{2} \cdot \frac{7}{8} = \boxed{\frac{7}{16}} \Rightarrow \textbf{(B)}</math>. | ||
− | ~AopsUser101 | + | ~AopsUser101 |
== Solution 3 == | == Solution 3 == |
Revision as of 15:46, 8 November 2022
Contents
[hide]Problem
Let and be the sequences of real numbers such that for all integers , where . What is
Solution 1
Square the given equality to yield so and
Solution 2 (DeMoivre's Formula)
Note that . Let , then, we know that so Therefore,
Aha! is a geometric sequence that evaluates to ! Now we can quickly see that Therefore, The imaginary part is , so our answer is .
~AopsUser101
Solution 3
Clearly . So we have . By linearity, we have the latter is equivalent to . Expanding the summand yields -vsamc
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=OdSTfCDOh5A&list=PLyhPcpM8aMvJvwA2kypmfdtlxH90ShZCc&index=2 - AMBRIGGS
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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.