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# Pre-RMO 2014/19

Let $$x_1, x_2, \ldots, x_{2014}$$ be real numbers different from $$1$$ such that $$x_1 + x_2 + \ldots + x_{2014} = 1$$ and

$\frac{x_1}{1 - x_1} + \frac{x_2}{1 - x_2} + \ldots + \frac{x_{2014}}{1 - x_{2014}} = 1.$

What is the value of

$\frac{x^2_1}{1 - x_1} + \frac{x^2_2}{1 - x_2} + \frac{x^2_3}{1 - x_3} + \ldots + \frac{x^2_{2014}}{1 - x_{2014}} ?$

This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
2 years, 9 months ago

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$$\displaystyle \frac{x_1^{2}}{1-x_1} + \frac{x_2^{2}}{1-x_2}+\ldots + \frac{x_{2014}^{2}}{1-x_{2014}}$$

$$\displaystyle x_1 + x_2 + \ldots + x_{2014}$$

To the above equation

$$\displaystyle \frac{x_1^{2}}{1-x_1} + x_1 + \frac{x_2^{2}}{1-x_2} + x_2 + \ldots \frac{x_{2014}^{2}}{1-x_{2014}} + x_{2014}$$

Now taking L.C.M

We finally get

$$\displaystyle \frac{x_1}{1-x_1} +\frac{x_2}{1-x_2} + \ldots \frac{x_{2014}}{1-x_{2014}}$$

So this is equal to 1 but we have added

$$x_1 + x_2 \ldots + x_{2014 }= 1$$

So we have to subtract 1

Finally we get the answer $$\boxed {0}$$ · 2 years, 9 months ago

Exactly did the same · 1 year, 2 months ago

Nice !!!!! · 2 years, 9 months ago

just subtract 1st eqn from 2..u get the required 3rd eqn =1-1=0 · 2 years, 7 months ago