Difference between revisions of "2019 AMC 12A Problems/Problem 21"
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The term <math>(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2})</math> simplifies to <math>6\mathrm{cis }(45)</math>, while the term <math>(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}})</math> simplifies to <math>\frac{6}{\mathrm{cis }(45)}</math>. Upon multiplication, the <math>\mathrm{cis }(45)</math> cancels out and leaves us with <math>\boxed{\textbf{(C) }36}</math>. | The term <math>(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2})</math> simplifies to <math>6\mathrm{cis }(45)</math>, while the term <math>(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}})</math> simplifies to <math>\frac{6}{\mathrm{cis }(45)}</math>. Upon multiplication, the <math>\mathrm{cis }(45)</math> cancels out and leaves us with <math>\boxed{\textbf{(C) }36}</math>. | ||
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+ | ==Solution 2== | ||
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+ | It is well known that if <math>|z|=1</math> then <math>\bar{z}=\frac{1}{z}</math>. Therefore, we have that the desired expression is equal to <cmath>(z^1+z^4+z^9+...+z^{144})(\bar{z}^1+\bar{z}^4+\bar{z}^9+...+\bar{z}^{144})</cmath> We know that <math>z=e^{\frac{i\pi}{4}}</math> so <math>\bar{z}=e^{\frac{i7\pi}{4}}</math>. Then, by De Moivre's Theorem, we have <cmath>(e^{\frac{i\pi}{4}}+e^{i\pi}+...+e^{2i\pi})(e^{\frac{i7\pi}{4}}+e^{i7\pi}+...+e^{2i\pi})</cmath> which can easily be computed as <math>\boxed{36}</math>. <math>\blacksquare</math> | ||
==See Also== | ==See Also== |
Revision as of 16:12, 14 February 2019
Contents
[hide]Problem
Let What is
Solution
Note that .
Also note that for all positive integers because of DeMoivre's Theorem. Therefore, we want to look at the exponents of each term mod 8.
and are all
and are all
and are all
and are all
Therefore,
The term simplifies to , while the term simplifies to . Upon multiplication, the cancels out and leaves us with .
Solution 2
It is well known that if then . Therefore, we have that the desired expression is equal to We know that so . Then, by De Moivre's Theorem, we have which can easily be computed as .
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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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