1994 AIME Problems/Problem 13

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 1

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 $9$ .

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

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 $\boxed{850}$ .

Solution 2

Divide both sides by $x^{10}$ to get $1 + \left(13-\dfrac{1}{x}\right)^{10}=0$

Rearranging: $\left(13-\dfrac{1}{x}\right)^{10} = -1$

Thus, $13-\dfrac{1}{x} = \omega$ where $\omega = e^{i(\pi n/5+\pi/10)}$ where $n$ is an integer.

We see that $\dfrac{1}{x}=13-\omega$ . Thus, $\dfrac{1}{x\overline{x}}=(13\, -\, \omega)(13\, -\, \overline{\omega})=169-13(\omega\, +\, \overline{\omega})\, +\, \omega\overline{\omega}=170\, -\, 13(\omega\, +\, \overline{\omega})$

Summing over all terms: $\dfrac{1}{r_1\overline{r_1}}+\cdots + \dfrac{1}{r_5\overline{r_5}} = 5\cdot 170 - 13(e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)})$

However, note that $e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)}=0$ from drawing the numbers on the complex plane, our answer is just $5\cdot 170=\boxed{850}$

Solution 3(Ileytyn)

Let us apply difference of squares, by writing this equation as $(x^5)^2-(i[13x-1]^5)^2=0$ , where $i=\sqrt{-1}$ .

Once applied, we have $(x^5-i[13x-1]^5)(x^5+i[13x-1]^5)=0$ By factorization, we get that either $\text{equation 1: }x^5-i[13x-1]^5=0\text{ or equation 2: }x^5+i[13x-1]^5=0$ We find the trivial solution to the first equation $x^5=i[13x-1]^5$ , and since a fifth root of $i=i^5$ is $i$ , we can find this solution by taking the fifth root, or when $x=(13x-1)i$ , which when solved, gives $x=\frac{i}{13i-1}=\frac{13-i}{170}$ . By using a similar approach to the second equation, $x^5+i[13x-1]^5=0$ , or $x=-i(13x-1)$ , we get $\frac{i}{13i+1}=\frac{13+i}{170}$ . We observe that these are conjugates. We also note that $z\overline{z}=|z|^2$ . Hence, we find the magnitude of one of these two conjugates, which is $\frac{1}{\sqrt{170}}$ , or as we are finding the reciprocal of the square value of this, this contributes $170$ to the final answer. We claim that each of the roots of equation one is a conjugate of one of equation 2. We also notice that, by an application of DeMoivre's, each of the roots has the same magnitude, just a different angle. Hence, as each of the reciprocals of product of conjugates has the value $170$ , and there are 5 of these pairs of conjugates, the answer is $170\cdot 5 = \boxed{850}$

Video Solution

https://youtu.be/3GG6tdEz0KA

1994 AIME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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