AoPS - Problem

WakeUp wrote: The real numbers

$x_1,\ldots ,x_{2011}$ satisfy

$x_1+x_2=2x_1',\ x_2+x_3=2x_2', \ \ldots, \ x_{2011}+x_1=2x_{2011}'$ where

$x_1',x_2',\ldots,x_{2011}'$ is a permutation of

$x_1,x_2,\ldots,x_{2011}$ . Prove that

$x_1=x_2=\ldots =x_{2011}$ . Chose any

$n<2011$ such equations : the number of

$x_i$ involved in these equations is clearly

$>n$ Let then

$M=\max(x_i)$ and let

$n$ be the number of

$x_i$ (and also of

$x_i'$ ) equal to

$M$ In each equality

$x_k+x_{k+1}=2M$ , we can deduce

$x_k=x_{k+1}=M$ Suppose

$n<2011$ : adding then the

$n$ such equalities, we get that the number of

$x_i=M$ is

$>n$ , and so contradiction. So

$n=2011$ Hence the result.