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1994 AIME Problems/Problem 13

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Problem

The equation

x^{10}+(13x-1)^{10}=0\,

has 10 complex roots r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\, where the bar denotes complex conjugation. Find the value of

\frac 1{r_1\overline{r_1}}+\frac 1{r_2\overline{r_2}}+\frac 1{r_3\overline{r_3}}+\frac 1{r_4\overline{r_4}}+\frac 1{r_5\overl...

Solution

Let t = 1/x. After multiplying the equation by t^{10}, 1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1.

Using DeMoivre, 13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) where k is an integer between 0 and 9.

t = 13 - \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) \Rightarrow \bar{t} = 13 - \text{cis}\left(-\frac {(2k + 1)\pi}{10}\r....

Since \text{cis}(\theta) + \text{cis}(-\theta) = 2\cos(\theta), t\bar{t} = 170 - 26\cos \left(\frac {(2k + 1)\pi}{10}\right) after expanding. Here k ranges from 0 to 4 because two angles which sum to 2\pi are involved in the product.

The expression to find is \sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}.

But \cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0 so the sum is \boxed{850}.

See also

1994 AIME (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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