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.

Use $\sum_{i}x_{i}^{2}=\sum_{i}x_{i}^{'2}=\sum_{i}\left(\frac{x_{i}+x_{i+1}}{2}\right)^{2} $ which can also written as $\sum_{i}\left(\frac{x_{i}-x_{i+1}}{2}\right)^{2}=0$