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

Revision as of 22:48, 4 July 2013

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $\boxed{784}$ .

@@ Line 14: / Line 14: @@
 [[Category:Intermediate Algebra Problems]]
 [[Category:Complex Number Problems]]
+{{MAA Notice}}

2011 AIME II (Problems • Answer Key • Resources)
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

Page

Toolbox

Search

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

Revision as of 22:48, 4 July 2013

Problem

Solution

See also