# 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.

• I am understanding your approach, give me time to understand it fully. So that I get my desired result.
– L.K.
Commented Jun 15, 2014 at 6:50
• If you tell me which part you don't understand, I may be able to give you more details. Commented Jun 16, 2014 at 4:43
• "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". From this you will show that $z_i$ are equal. Can You explain me the text in "".
– L.K.
Commented Jun 16, 2014 at 5:11
• Each $z_i\leq t$ because $t$ was the maximum value. So for each blue dot $z_i'$ we have $2z_i'=z_i+z_{i+1}\leq 2t$, so $z_i'\leq t$. $z_i'$ can only be equal to $t$ if both $z_i$ and $z_{i+1}$ are equal to $t$. Commented Jun 16, 2014 at 6:30