1994 AIME Problem 13

Question

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

\begin{aligned}&\frac1{r_1\overline{r_1}}+\frac1{r_2\overline{r_2}}+\frac1{r_3\overline{r_3}}\\&\quad+\frac1{r_4\overline{r_4}}+\frac1{r_5\overline{r_5}}.\end{aligned}

Answer

Set

\frac{x}{13x-1}=\zeta,

where

\zeta^{10}=-1

Express

\frac{1}{|x|^2}

in terms of

\zeta+\overline\zeta

and sum over the five conjugate pairs

Let

\zeta=\frac{x}{13x-1},

so

\zeta^{10}=-1

and

\begin{aligned}x&=\frac{\zeta}{13\zeta-1},\\\frac1x&=13-\zeta^{-1}.\end{aligned}

Since

|\zeta|=1,

\begin{aligned}\frac1{|x|^2}&=|13-\zeta^{-1}|^2\\&=170-13(\zeta+\overline\zeta).\end{aligned}

Summing one value for each of the five conjugate pairs gives

5\cdot170

minus

13

times the sum of all ten roots of

z^{10}+1,

which is

0.

The requested value is

850.

Problem 13 in other years