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Difference between revisions of "2000 AIME II Problems/Problem 9"

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== Problem ==
 
== Problem ==
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The system of equations
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Given that <math>z</math> is a complex number such that <math>z+\frac 1z=2\cos 3^\circ</math>, find the least integer that is greater than <math>z^{2000}+\frac 1{z^{2000}}</math>.
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<center><math>\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\
 
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\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\
 
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\log_{10}(2000zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\
 
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\end{eqnarray*}</math></center>
 
  
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has two solutions <math>(x_{1},y_{1},z_{1})</math> and <math>(x_{2},y_{2},z_{2})</math>. Find <math>y_{1} + y_{2}</math>.
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== Solution ==
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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>.
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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>.
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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
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<math>z^{2000} = \cos 240^\circ + i\sin 240^\circ</math>.
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We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math>.
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Finally, the least integer greater than <math>-1</math> is <math>\boxed{000}</math>.
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==Solution 2==
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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>
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<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>
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 +
~solution by williamgolly
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== Solution 3 Intuitive ==
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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.
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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>.
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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>
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~ PaperMath ~megaboy6679
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== Solution 4 ==
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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.
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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>
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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>.
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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.  
  
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== Solution ==
 
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{{solution}}
 
  
 
== 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}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:24, 2 November 2024

Problem

Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$, find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$.

Solution

Using the quadratic equation on $z^2 - (2 \cos 3 )z + 1 = 0$, we have $z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}$.

There are other ways we can come to this conclusion. Note that if $z$ is on the unit circle in the complex plane, then $z = e^{i\theta} = \cos \theta + i\sin \theta$ and $\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta$. We have $z+\frac 1z = 2\cos \theta = 2\cos 3^\circ$ and $\theta = 3^\circ$. Alternatively, we could let $z = a + bi$ and solve to get $z=\cos 3^\circ + i\sin 3^\circ$.


Using De Moivre's Theorem we have $z^{2000} = \cos 6000^\circ + i\sin 6000^\circ$, $6000 = 16(360) + 240$, so $z^{2000} = \cos 240^\circ + i\sin 240^\circ$.

We want $z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1$.

Finally, the least integer greater than $-1$ is $\boxed{000}$.

Solution 2

Let $z=re^{i\theta}$. Notice that we have $2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.$

$r$ must be $1$ (or if you take the magnitude would not be the same). Therefore, $z=e^{i\frac{\pi}{\theta}}$ and plugging into the desired expression, we get $e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1$. Therefore, the least integer greater is $\boxed{000}.$

~solution by williamgolly


Solution 3 Intuitive

For this solution, we assume that $z^{2000} + 1/z^{2000}$ and $z^{2048} + 1/z^{2048}$ has the same least integer greater than their solution.

We have $z + 1/z = 2\cos 3$. Since $\cos 3<1$, $2\cos 3<2$. If we square the equation $z + 1/z = 2\cos 3$, we get $z^2 + 2 + 1/(z^2) = 4\cos^2 3$, or $z^2 + 1/(z^2) = 4\cos^2 3 - 2$. $4\cos^2 3 - 2$ is is less than $2$, since $4\cos^2 3$ is less than $4$. If we square the equation again, we get $z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2$.

Since $4\cos^2 3 - 2$ is less than 2, $(4\cos^2 3 - 2)^2$ is less than 4, and $(4\cos^2 3 - 2)^2 -2$ is less than 2. However $(4\cos^2 3 - 2)^2 -2$ is also less than $4\cos^2 3 - 2$. 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 $\boxed{000}.$

~ PaperMath ~megaboy6679

Solution 4

First, let $z = a+bi$ where $a$ and $b$ are real numbers. We now have that \[a+bi+\frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}\] given the conditions of the problem. Equating imaginary coefficients, we have that \[b \left( 1 - \frac{1}{a^2+b^2}\right) = 0\] giving us that either $b=0$ or $|z| = 1$. Let's consider the latter case for now.

We now know that $a^2+b^2=1$, so when we equate real coefficients we have that $2a = 2 \cos{3^{\circ}}$, therefore $a = \cos{3^{\circ}}$. So, $b = \cos{3^{\circ}}$ and then we can write $z = \text{cis}(3)^{\circ}$

By De Moivre's Theorem, \[z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}\]. The imaginary parts cancel, leaving us with $2 \cos{6000^{\circ}}$, which is $240 \pmod{360}$. Therefore, it is $-1$, and our answer is $\boxed{000}$.

Now, if $b=0$ then we have that $a+\frac{1}{a} = 2 \cos{3^{\circ}}$. Therefore, $a$ is not violating our conditions set above.


See also

2000 AIME II (Problems • Answer Key • Resources)
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Problem 8 		Followed by
Problem 10
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