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>159 + 323 = \ | + | <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 19:02, 4 July 2013
Problem
The polynomial 
has 
complex roots of the form ![$z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34,$](http://latex.artofproblemsolving.com/d/0/c/d0c96625746cab4901eb36724577280cf1515b7b.png)
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. 
