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Difference between revisions of "2011 AIME II Problems/Problem 8"
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== Problem ==
  
 
Let <math>z_1</math>, <math>z_2</math>, <math>z_3</math>, <math>\dots</math>, <math>z_{12}</math> be the 12 zeroes of the polynomial <math>z^{12} - 2^{36}</math>. For each <math>j</math>, let <math>w_j</math> be one of <math>z_j</math> or <math>iz_j</math>. Then the maximum possible value of the real part of <math>\sum_{j = 1}^{12} w_j</math> can be written as <math>m + \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m + n</math>.
 
Let <math>z_1</math>, <math>z_2</math>, <math>z_3</math>, <math>\dots</math>, <math>z_{12}</math> be the 12 zeroes of the polynomial <math>z^{12} - 2^{36}</math>. For each <math>j</math>, let <math>w_j</math> be one of <math>z_j</math> or <math>iz_j</math>. Then the maximum possible value of the real part of <math>\sum_{j = 1}^{12} w_j</math> can be written as <math>m + \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m + n</math>.
  
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== Solution ==
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Solution:
 
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Let me first note that this is a hastily compiled solution.
 
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<geogebra>973797c41824115b678d18138dff008b8acfda12</geogebra>
 
  
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If you multiply one of those dots on the circle(of radius 8) by i, you move it 90 degrees clockwise.  You want everything to be as far right as possible, and when you are done, you find that you get '''784''' when you add everything.
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[[File:2011_AIME_II_-8.png‎]]
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The twelve dots above represent the 12 roots of the equation <math>z^{12}-2^{36}=0</math>. If we write <math>z=a+bi</math>, then the real part of <math>z</math> is <math>a</math> and the real part of <math>iz</math> is <math>-b</math>. The blue dots represent those roots <math>z</math> for which the real part of <math>z</math> is greater than the real part of <math>iz</math>, and the red dots represent those roots <math>z</math> for which the real part of <math>iz</math> is greater than the real part of <math>z</math>. Now, the sum of the real parts of the blue dots is easily seen to be <math>8+16\cos\frac{\pi}{6}=8+8\sqrt{3}</math> and the negative of the sum of the imaginary parts of the red dots is easily seen to also be <math>8+8\sqrt{3}</math>. Hence our desired sum is <math>16+16\sqrt{3}=16+\sqrt{768}</math>, giving the answer <math>\boxed{784}</math>.

Revision as of 16:36, 3 April 2011

Problem

Let $z_1$, $z_2$, $z_3$, $\dots$, $z_{12}$ be the 12 zeroes of the polynomial $z^{12} - 2^{36}$. For each $j$, let $w_j$ be one of $z_j$ or $iz_j$. Then the maximum possible value of the real part of $\sum_{j = 1}^{12} w_j$ can be written as $m + \sqrt{n}$, where $m$ and $n$ are positive integers. Find $m + n$.

Solution

2011 AIME II -8.png

The twelve dots above represent the 12 roots of the equation $z^{12}-2^{36}=0$. If we write $z=a+bi$, then the real part of $z$ is $a$ and the real part of $iz$ is $-b$. The blue dots represent those roots $z$ for which the real part of $z$ is greater than the real part of $iz$, and the red dots represent those roots $z$ for which the real part of $iz$ is greater than the real part of $z$. Now, the sum of the real parts of the blue dots is easily seen to be $8+16\cos\frac{\pi}{6}=8+8\sqrt{3}$ and the negative of the sum of the imaginary parts of the red dots is easily seen to also be $8+8\sqrt{3}$. Hence our desired sum is $16+16\sqrt{3}=16+\sqrt{768}$, giving the answer $\boxed{784}$.

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