RMO 2015!! #5

Let $\omega$ denote complex(non-real) root of unity.

Since $x,y \neq 0$ . So,

$\left(\dfrac{x}{y} \right)^{2} + \dfrac{x}{y} + 1 = 0$

So, $\dfrac{x}{y} = \omega$

So, $\left(\dfrac{x}{x+y}\right)^{2001} + \left(\dfrac{y}{x+y}\right)^{2001} =\left(\dfrac{1}{1+\dfrac{1}{\omega}}\right)^{2001} \left(\dfrac{1}{1+\omega}\right)^{2001}$ = $(-\omega)^{2001} + (-\omega ^2)^{2001}$ = $\boxed{-2}$ .

Simply,

$x^2 + xy + y^2=(x-\omega y)(x-\omega^2y)\\ \begin{array}{l||c|r|}\Rightarrow&(x-\omega y)=0&(x-\omega^2y)=0\\\hline x&\omega y&\omega^2y\\y&\omega^2x&\omega x\end{array}\\ \begin{aligned}\text{Taking any one of the value, } \dfrac{x}{x+y} &= \dfrac{x}{(1+\omega^2)x}= -\omega^2\\ \text{Similarly, }\dfrac{y}{x+y} &= \dfrac{y}{(1+\omega)y}= -\omega\end{aligned}$ $\begin{aligned}\Rightarrow \left( \frac{x}{ x+y } \right)^{2001} + \left( \frac{y}{x+y}\right) ^ { 2001}&= -\left[\omega^{2001} + \left(\omega^2\right)^{2001}\right] \\&=-2\end{aligned}$

$\Large\boxed{\left( \frac{x}{ x+y } \right)^{2001} + \left( \frac{y}{x+y}\right) ^ { 2001} = -2}$

The equation can be written as x/y + y/x = -1 Put x/y = t, then t + 1/t = -1 OR t + 1/t +1 = 0. The roots of the given equation are w, w^2 i.e. x / y = w or w^2. Clearly x= yw or yw^2.

Substitute the obtained value in the equation and simplify to obtain the answer.