Difference between revisions of "1994 AIME Problems/Problem 13"
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The equation | The equation | ||
<center><math>x^{10}+(13x-1)^{10}=0\,</math></center> | <center><math>x^{10}+(13x-1)^{10}=0\,</math></center> | ||
− | has 10 complex | + | has 10 [[complex number|complex]] [[root]]s <math>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},\,</math> where the bar denotes complex conjugation. Find the value of |
<center><math>\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}}.</math></center> | <center><math>\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}}.</math></center> | ||
− | == Solution == | + | == Solution 1 == |
− | {{ | + | 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>. |
+ | |||
+ | Using DeMoivre, <math>13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right)</math> where <math>k</math> is an integer between <math>0</math> and <math>9</math>. | ||
+ | |||
+ | <math>t = 13 - \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) \Rightarrow \bar{t} = 13 - \text{cis}\left(-\frac {(2k + 1)\pi}{10}\right)</math>. | ||
+ | |||
+ | Since <math>\text{cis}(\theta) + \text{cis}(-\theta) = 2\cos(\theta)</math>, <math>t\bar{t} = 170 - 26\cos \left(\frac {(2k + 1)\pi}{10}\right)</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. | ||
+ | |||
+ | The expression to find is <math>\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Divide both sides by <math>x^{10}</math> to get <cmath>1 + \left(13-\dfrac{1}{x}\right)^{10}=0</cmath> | ||
+ | |||
+ | Rearranging: <cmath>\left(13-\dfrac{1}{x}\right)^{10} = -1</cmath> | ||
+ | |||
+ | Thus, <math>13-\dfrac{1}{x} = \omega</math> where <math>\omega = e^{i(\pi n/5+\pi/10)}</math> where <math>n</math> is an integer. | ||
+ | |||
+ | We see that <math>\dfrac{1}{x}=13-\omega</math>. Thus, <cmath>\dfrac{1}{x\overline{x}}=(13-\omega)(13-\overline{\omega})=169-13(\omega+\overline{\omega})+\omega\overline{\omega}=170-13(\omega+\overline{\omega})</cmath> | ||
+ | |||
+ | Summing over all terms: <cmath>\dfrac{1}{r_1\overline{r_1}}+\cdots + \dfrac{1}{r_5\overline{r_5}} = 5\cdot 170 - 13(e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)})</cmath> | ||
+ | |||
+ | However, note that <math>e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)}=0</math> from drawing the numbers on the complex plane, our answer is just <cmath>5\cdot 170=\boxed{850}</cmath> | ||
== See also == | == See also == | ||
{{AIME box|year=1994|num-b=12|num-a=14}} | {{AIME box|year=1994|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:51, 6 December 2014
Contents
Problem
The equation
has 10 complex roots where the bar denotes complex conjugation. Find the value of
Solution 1
Let . After multiplying the equation by , .
Using DeMoivre, where is an integer between and .
.
Since , after expanding. Here ranges from 0 to 4 because two angles which sum to are involved in the product.
The expression to find is .
But so the sum is .
Solution 2
Divide both sides by to get
Rearranging:
Thus, where where is an integer.
We see that . Thus,
Summing over all terms:
However, note that from drawing the numbers on the complex plane, our answer is just
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.