Difference between revisions of "2011 AIME II Problems/Problem 8"

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== Problem ==
 
== 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} z_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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Let <math>z_1,z_2,z_3,\dots,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>i z_j</math>. Then the maximum possible value of the real part of <math>\displaystyle\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>.
  
 
== Solution ==
 
== Solution ==

Revision as of 20:57, 17 March 2019

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 $i z_j$. Then the maximum possible value of the real part of $\displaystyle\sum_{j=1}^{12} w_j$ 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 $\boxed{784}$.

See also

2011 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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