Difference between revisions of "2000 AIME II Problems/Problem 9"
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We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math>. | We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math>. | ||
− | Finally, the least integer greater than <math>-1</math> is <math>\boxed{ | + | Finally, the least integer greater than <math>-1</math> is <math>\boxed{000}</math>. |
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>z=re^{i\theta}</math>. Notice that we have <math>2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.</math> | ||
+ | |||
+ | <math>r</math> must be <math>1</math> (or else if you take the magnitude would not be the same). Therefore, <math>z=e^{i\frac{\pi}{\theta}}</math> and plugging into the desired expression, we get <math>e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1</math>. Therefore, the least integer greater is <math>\boxed{000}.</math> | ||
+ | |||
+ | ~solution by williamgolly | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 21:06, 15 July 2020
Contents
Problem
Given that is a complex number such that , find the least integer that is greater than .
Solution
Using the quadratic equation on , we have .
There are other ways we can come to this conclusion. Note that if is on the unit circle in the complex plane, then and . We have and . Alternatively, we could let and solve to get .
Using De Moivre's Theorem we have , , so
.
We want .
Finally, the least integer greater than is .
Solution 2
Let . Notice that we have
must be (or else if you take the magnitude would not be the same). Therefore, and plugging into the desired expression, we get . Therefore, the least integer greater is
~solution by williamgolly
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.