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

(Solution)
(Solution 2)
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== Solution 2 ==
 
== Solution 2 ==
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The equation <math>z^12-2^36</math> can be factored as follows:
  
 
==See also==
 
==See also==

Revision as of 22:08, 11 December 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 $\boxed{784}$.

Solution 2

The equation $z^12-2^36$ can be factored as follows:

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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