Not able to solve this algebra problem. I tried it, but didn't get anywhere:
The real numbers $z_1,\dots ,z_{2011} $ satisfy
$z_1 + z_2 = 2z'_1 ,\hspace{1cm} z_2 + z_3 = 2z'_2 ,\hspace{0.5cm} \dots , z_{2011} + z_1 = 2z'_{2011}$
where $z'_1, z'_2, \dots, z'_{2011}$ is a permutation of $z_1,\dots ,z_{2011}$. Prove that $z_1 = z_2 = \cdots = z_{2011}$.

Let $t$ be the maximum value attained by any of the $z_i$
We arrange the $z_i$ on a circle as red dots and the $z'_i$ inbetween $z_i$ and $z_{i+1}$ as blue dots.
So every $z_i$ is really is twice on the circle, once as a blue dot and once as a red dot.
This specifically means that there are as many red dots as blue dots with value $t$.
By definition the value of the blue dots is the average of the value of its two neighbours.
Because $t$ was the maximum value, every blue dot with value $t$ must have two neighbours of value $t$.
If every dot has value $t$ we are done.
Otherwise we walk around the circle and gather the maximal lines where each dot
has value $t$. We just argued that the endpoints of such lines must all be red.
But this implies that there are more red dots then blue dots with value $t$, which
is a contradiction.
