Difference between revisions of "2004 AIME I Problems/Problem 13"
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== Problem == | == Problem == | ||
− | The polynomial <math> P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17} </math> has 34 complex roots of the form <math> z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34, </math> with <math> 0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{34} < 1 </math> and <math> r_k>0. </math> Given that <math> a_1 + a_2 + a_3 + a_4 + a_5 = m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math> | + | The polynomial <math> P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17} </math> has <math>34</math> complex roots of the form <math> z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34, </math> with <math> 0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{34} < 1 </math> and <math> r_k>0. </math> Given that <math> a_1 + a_2 + a_3 + a_4 + a_5 = m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math> |
== Solution == | == Solution == | ||
We see that the expression for the [[polynomial]] <math>P</math> is very difficult to work with directly, but there is one obvious transformation to make: sum the [[geometric series]]: | We see that the expression for the [[polynomial]] <math>P</math> is very difficult to work with directly, but there is one obvious transformation to make: sum the [[geometric series]]: | ||
− | < | + | <cmath>\begin{align*} |
+ | P(x) &= \left(\frac{x^{18} - 1}{x - 1}\right)^2 - x^{17} = \frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17}\\ &= \frac{x^{36} - x^{19} - x^{17} + 1}{(x - 1)^2} = \frac{(x^{19} - 1)(x^{17} - 1)}{(x - 1)^2} \end{align*}</cmath> | ||
− | This [[expression]] has [[root]]s at every | + | This [[expression]] has [[root]]s at every <math>17</math>th root and <math>19</math>th [[roots of unity]], other than <math>1</math>. Since <math>17</math> and <math>19</math> are [[relatively prime]], this means there are no duplicate roots. Thus, <math>a_1, a_2, a_3, a_4</math> and <math>a_5</math> are the five smallest fractions of the form <math>\frac m{19}</math> or <math>\frac n {17}</math> for <math>m, n > 0</math>. |
<math>\frac 3 {17}</math> and <math>\frac 4{19}</math> can both be seen to be larger than any of <math>\frac1{19}, \frac2{19}, \frac3{19}, \frac 1{17}, \frac2{17}</math>, so these latter five are the numbers we want to add. | <math>\frac 3 {17}</math> and <math>\frac 4{19}</math> can both be seen to be larger than any of <math>\frac1{19}, \frac2{19}, \frac3{19}, \frac 1{17}, \frac2{17}</math>, so these latter five are the numbers we want to add. | ||
− | <math>\frac1{19}+ \frac2{19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}</math> and so the answer is <math> | + | <math>\frac1{19}+ \frac2{19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}</math> and so the answer is <math>159 + 323 = \boxed{482}</math>. |
== See also == | == See also == | ||
− | + | {{AIME box|year=2004|n=I|num-b=12|num-a=14}} | |
− | + | [[Category:Intermediate Algebra Problems]] | |
− | + | {{MAA Notice}} | |
− |
Latest revision as of 20:02, 4 July 2013
Problem
The polynomial has complex roots of the form with and Given that where and are relatively prime positive integers, find
Solution
We see that the expression for the polynomial is very difficult to work with directly, but there is one obvious transformation to make: sum the geometric series:
This expression has roots at every th root and th roots of unity, other than . Since and are relatively prime, this means there are no duplicate roots. Thus, and are the five smallest fractions of the form or for .
and can both be seen to be larger than any of , so these latter five are the numbers we want to add.
and so the answer is .
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.