UkraineMO

Let $|a_0 - a_1| = k$. We can write down the equations:
$$\begin{align*} a_0 - a_1 &= \pm k, \\ a_1 - a_2 &= \pm \frac{2^0}{3^1} k, \\ a_2 - a_3 &= \pm \frac{2^1}{3^2} k, \\ \dots, \\ a_{2011} - a_{2012} &= \pm \frac{2^{2010}}{3^{2011}} k, \\ a_{2012} - a_0 &= \pm \frac{2^{2011}}{3^{2012}} k. \end{align*}$$
If $k \neq 0$, let us add these equations and reduce everything by $k$, as a result, we will get an equation:
$$0 = \pm 1 \pm \frac{2^0}{3^1} \pm \frac{2^1}{3^2} \pm \dots \pm \frac{2^{2010}}{3^{2011}} \pm \frac{2^{2011}}{3^{2012}}.$$
But irrespectively of sign before the 1 we cannot obtain 0 as a result, because module of sum of the summands left is less than 1. It is easy to prove using the formula for the sum of members of geometric progression:
$$\left| \pm \frac{2^0}{3^1} \pm \frac{2^1}{3^2} \pm \dots \pm \frac{2^{2010}}{3^{2011}} \pm \frac{2^{2011}}{3^{2012}} \right| \le \frac{2^0}{3^1} + \frac{2^1}{3^2} + \dots + \frac{2^{2010}}{3^{2011}} + \frac{2^{2011}}{3^{2012}} = \frac{\frac{1}{3}\left(1 - \left(\frac{2}{3}\right)^{2012}\right)}{1 - \frac{2}{3}} = 1 - \left(\frac{2}{3}\right)^{2012} < 1.$$