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

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== See also ==
 
== See also ==
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* [[2004 AIME I Problems/Problem 12| Previous problem]]
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* [[2004 AIME I Problems/Problem 14| Next problem]]
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* [[2004 AIME I Problems]]
 
* [[2004 AIME I Problems]]

Revision as of 02:45, 6 November 2006

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:

$P(x) = (\frac{x^{18} - 1}{x - 1})^2 - x^{17} = \frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17} = \frac{x^{36} - x^{19} - x^{17} + 1}{(x - 1)^2} = \frac{(x^{19} - 1)(x^{17} - 1)}{(x - 1)^2}$.

This expression has roots 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, $a_1, a_2, a_3, a_4$ 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 $\displaystyle 159 + 323 = 482$.

See also

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