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Difference between revisions of "2004 AIME I Problems/Problem 13"

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== Problem ==
 
== Problem ==
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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>
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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]]:
  
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<math>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}</math>.
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<cmath>\begin{align*}
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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>
  
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This [[expression]] has [[root]]s at every 17th root and 19th root of unity, other than 1.  Since 17 and 19 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>.   
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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.
  
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<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>\displaystyle 159 + 323 = 482</math>.
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<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 ==
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* [[2004 AIME I Problems/Problem 12| Previous problem]]
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{{AIME box|year=2004|n=I|num-b=12|num-a=14}}
  
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* [[2004 AIME I Problems/Problem 14| Next problem]]
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}
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* [[2004 AIME I Problems]]
 

Latest revision as of 19:02, 4 July 2013

Problem

The polynomial $P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}$ has $34$ complex roots of the form $z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34,$ with $0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{34} < 1$ and $r_k>0.$ Given that $a_1 + a_2 + a_3 + a_4 + a_5 = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

We see that the expression for the polynomial $P$ is very difficult to work with directly, but there is one obvious transformation to make: sum the geometric series:

\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*}

This expression has roots at every $17$th root and $19$th roots of unity, other than $1$. Since $17$ and $19$ are relatively prime, this means there are no duplicate roots. Thus, $a_1, a_2, a_3, a_4$ and $a_5$ are the five smallest fractions of the form $\frac m{19}$ or $\frac n {17}$ for $m, n > 0$.

$\frac 3 {17}$ and $\frac 4{19}$ can both be seen to be larger than any of $\frac1{19}, \frac2{19}, \frac3{19}, \frac 1{17}, \frac2{17}$, so these latter five are the numbers we want to add.

$\frac1{19}+ \frac2{19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}$ and so the answer is $159 + 323 = \boxed{482}$.

See also

2004 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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