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} 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>. |
== Solution == | == Solution == | ||
+ | <center>[[File:2011_AIME_II_-8.png]]</center> | ||
+ | {{image}} <!-- convert to asymptote! --> | ||
− | + | 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>. | |
− | + | ==See also== | |
+ | {{AIME box|year=2011|n=II|num-b=7|num-a=9}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | [[Category:Complex Number Problems]] |
Revision as of 09:41, 23 August 2011
Problem
Let , , , , be the 12 zeroes of the polynomial . For each , let be one of or . Then the maximum possible value of the real part of can be written as , where and are positive integers. Find .
Solution
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The twelve dots above represent the 12 roots of the equation . If we write , then the real part of is and the real part of is . The blue dots represent those roots for which the real part of is greater than the real part of , and the red dots represent those roots for which the real part of is greater than the real part of . Now, the sum of the real parts of the blue dots is easily seen to be and the negative of the sum of the imaginary parts of the red dots is easily seen to also be . Hence our desired sum is , giving the answer .
See also
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 |