Difference between revisions of "2000 AIME II Problems/Problem 9"
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== Solution == | == Solution == | ||
− | + | Using the quadratic equation on <math>z^2 - (2 \cos 3 )z + 1 = 0</math>, we have <math>z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}</math>. | |
− | We have <math>z+\frac 1z = 2\cos \theta = 2\cos 3^\circ</math> and <math>\theta = 3^\circ</math> | + | There are other ways we can come to this conclusion. Note that if <math>z</math> is on the [[unit circle]] in the complex plane, then <math>z = e^{i\theta} = \cos \theta + i\sin \theta</math> and <math>\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta</math>. We have <math>z+\frac 1z = 2\cos \theta = 2\cos 3^\circ</math> and <math>\theta = 3^\circ</math>. Alternatively, we could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math>. |
− | Alternatively, we could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math> | ||
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− | We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math> | + | Using [[De Moivre's Theorem]] we have <math>z^{2000} = \cos 6000^\circ + i\sin 6000^\circ</math>, <math>6000 = 16(360) + 240</math>, so |
+ | <math>z^{2000} = \cos 240^\circ + i\sin 240^\circ</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{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 == | ||
{{AIME box|year=2000|n=II|num-b=8|num-a=10}} | {{AIME box|year=2000|n=II|num-b=8|num-a=10}} | ||
+ | |||
+ | [[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.