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Difference between revisions of "1994 AIME Problems/Problem 13"

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== Solution ==
 
== Solution ==
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{{solution}}
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Let <math>t = 1/x</math>. After multiplying the equation by <math>t^{10}</math>, <math>1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1</math>.
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Using DeMoivre, <math>13 - t = e^\frac {(2k + 1)\pi}{10}</math> where <math>k</math> is an integer between <math>0</math> and <math>9</math>.
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<math>t = 13 - e^\frac {(2k + 1)\pi}{10} \Rightarrow \bar{t} = 13 - e^{ - \frac {(2k + 1)\pi}{10}}</math>.
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Since <math>e^{iy} + e^{ - iy} = 2\cos y</math>, <math>t\bar{t} = 170 - 2\cos \frac {(2k + 1)\pi}{10}</math> after expanding. Here <math>k</math> ranges from 0 to 4 because two angles which sum to <math>2\pi</math> are involved in the product..
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The expression to find is <math>\sum t\bar{t} = 850 - 2\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}</math>.
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But <math>\cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0</math> so the sum is <math>\boxed{850}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:46, 27 November 2008

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

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

Solution

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

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

$t = 13 - e^\frac {(2k + 1)\pi}{10} \Rightarrow \bar{t} = 13 - e^{ - \frac {(2k + 1)\pi}{10}}$.

Since $e^{iy} + e^{ - iy} = 2\cos y$, $t\bar{t} = 170 - 2\cos \frac {(2k + 1)\pi}{10}$ 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 $\sum t\bar{t} = 850 - 2\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}$.

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 (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
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