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> | ||
− | |||
− | |||
− | 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 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 | ||
+ | |||
+ | |||
+ | == Solution 3 Intuitive == | ||
+ | |||
+ | For this solution, we assume that <math>z^{2000} + 1/z^{2000}</math> and <math>z^{2048} + 1/z^{2048}</math> has the same least integer greater than their solution. | ||
+ | |||
+ | We have <math>z + 1/z = 2\cos 3</math>. Since <math>\cos 3<1</math>, <math>2\cos 3<2</math>. If we square the equation <math>z + 1/z = 2\cos 3</math>, we get <math>z^2 + 2 + 1/(z^2) = 4\cos^2 3</math>, or <math>z^2 + 1/(z^2) = 4\cos^2 3 - 2</math>. <math>4\cos^2 3 - 2</math> is is less than <math>2</math>, since <math>4\cos^2 3</math> is less than <math>4</math>. If we square the equation again, we get <math>z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2</math>. | ||
+ | |||
+ | Since <math>4\cos^2 3 - 2</math> is less than 2, <math>(4\cos^2 3 - 2)^2</math> is less than 4, and <math>(4\cos^2 3 - 2)^2 -2</math> is less than 2. However <math>(4\cos^2 3 - 2)^2 -2</math> is also less than <math>4\cos^2 3 - 2</math>. we can see that every time we square the equation, the right-hand side gets smaller and into the negatives. Since the smallest integer that is allowed as an answer is 0, the smallest integer greater is <math>\boxed{000}.</math> | ||
+ | |||
+ | ~ PaperMath ~megaboy6679 | ||
+ | |||
+ | == Solution 4 == | ||
+ | First, let <math>z = a+bi</math> where <math>a</math> and <math>b</math> are real numbers. We now have that <cmath>a+bi+\frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}</cmath> given the conditions of the problem. Equating imaginary coefficients, we have that <cmath>b \left( 1 - \frac{1}{a^2+b^2}\right) = 0</cmath> giving us that either <math>b=0</math> or <math>|z| = 1</math>. Let's consider the latter case for now. | ||
+ | |||
+ | We now know that <math>a^2+b^2=1</math>, so when we equate real coefficients we have that <math>2a = 2 \cos{3^{\circ}}</math>, therefore <math>a = \cos{3^{\circ}}</math>. So, <math>b = \cos{3^{\circ}}</math> and then we can write <math>z = \text{cis}(3)^{\circ}</math> | ||
+ | |||
+ | By De Moivre's Theorem, <cmath>z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}</cmath>. The imaginary parts cancel, leaving us with <math>2 \cos{6000^{\circ}}</math>, which is <math>240 \pmod{360}</math>. Therefore, it is <math>-1</math>, and our answer is <math>\boxed{000}</math>. | ||
+ | |||
+ | Now, if <math>b=0</math> then we have that <math>a+\frac{1}{a} = 2 \cos{3^{\circ}}</math>. Therefore, <math>a</math> is not violating our conditions set above. | ||
− | |||
== 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}} |
Latest revision as of 19:24, 2 November 2024
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 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
Solution 3 Intuitive
For this solution, we assume that and has the same least integer greater than their solution.
We have . Since , . If we square the equation , we get , or . is is less than , since is less than . If we square the equation again, we get .
Since is less than 2, is less than 4, and is less than 2. However is also less than . we can see that every time we square the equation, the right-hand side gets smaller and into the negatives. Since the smallest integer that is allowed as an answer is 0, the smallest integer greater is
~ PaperMath ~megaboy6679
Solution 4
First, let where and are real numbers. We now have that given the conditions of the problem. Equating imaginary coefficients, we have that giving us that either or . Let's consider the latter case for now.
We now know that , so when we equate real coefficients we have that , therefore . So, and then we can write
By De Moivre's Theorem, . The imaginary parts cancel, leaving us with , which is . Therefore, it is , and our answer is .
Now, if then we have that . Therefore, is not violating our conditions set above.
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 |
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