Difference between revisions of "2019 AMC 12A Problems/Problem 21"
Shankarmath (talk | contribs) (→Solution) |
Sevenoptimus (talk | contribs) m (Fixed formatting (and added degree symbols)) |
||
Line 6: | Line 6: | ||
==Solution 1== | ==Solution 1== | ||
− | Note that <math>z = \mathrm{cis }(45)</math>. | + | Note that <math>z = \mathrm{cis }(45^{\circ})</math>. |
− | Also note that <math>z^{k} = z^{k + 8}</math> for all positive integers <math>k</math> because of | + | Also note that <math>z^{k} = z^{k + 8}</math> for all positive integers <math>k</math> because of De Moivre's Theorem. Therefore, we want to look at the exponents of each term modulo <math>8</math>. |
<math>1^2, 5^2,</math> and <math>9^2</math> are all <math>1 \pmod{8}</math> | <math>1^2, 5^2,</math> and <math>9^2</math> are all <math>1 \pmod{8}</math> | ||
Line 20: | Line 20: | ||
Therefore, | Therefore, | ||
− | <math>z^{1^2} = z^{5^2} = z^{9^2} = \mathrm{cis }(45)</math> | + | <math>z^{1^2} = z^{5^2} = z^{9^2} = \mathrm{cis }(45^{\circ})</math> |
− | <math>z^{2^2} = z^{6^2} = z^{10^2} = \mathrm{cis }(180) = -1</math> | + | <math>z^{2^2} = z^{6^2} = z^{10^2} = \mathrm{cis }(180^{\circ}) = -1</math> |
− | <math>z^{3^2} = z^{7^2} = z^{11^2} = \mathrm{cis }(45)</math> | + | <math>z^{3^2} = z^{7^2} = z^{11^2} = \mathrm{cis }(45^{\circ})</math> |
− | <math>z^{4^2} = z^{8^2} = z^{12^2} = \mathrm{cis }(0) = 1</math> | + | <math>z^{4^2} = z^{8^2} = z^{12^2} = \mathrm{cis }(0^{\circ}) = 1</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>. | + | The term thus <math>(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2})</math> simplifies to <math>6\mathrm{cis }(45^{\circ})</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^{\circ})}</math>. Upon multiplication, the <math>\mathrm{cis }(45^{\circ})</math> cancels out and leaves us with <math>\boxed{\textbf{(C) }36}</math>. |
==Solution 2== | ==Solution 2== | ||
− | 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>. | + | 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>. |
==See Also== | ==See Also== |
Revision as of 21:03, 17 February 2019
Contents
[hide]Problem
Let What is
Solution 1
Note that .
Also note that for all positive integers because of De Moivre's Theorem. Therefore, we want to look at the exponents of each term modulo .
and are all
and are all
and are all
and are all
Therefore,
The term thus 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.