Question 71
The real numbers $x_1, \dots, x_{2011}$ satisfy
$$
x_1 + x_2 = 2x'_1, \quad x_2 + x_3 = 2x'_2, \quad \dots, \quad x_{2011} + x_1 = 2x'_{2011}
$$
where $x'_1, x'_2, \dots, x'_{2011}$ is a permutation of $x_1, x_2, \dots, x_{2011}$. Prove $x_1 = x_2 = \dots = x_{2011}$.
$$
x_1 + x_2 = 2x'_1, \quad x_2 + x_3 = 2x'_2, \quad \dots, \quad x_{2011} + x_1 = 2x'_{2011}
$$
where $x'_1, x'_2, \dots, x'_{2011}$ is a permutation of $x_1, x_2, \dots, x_{2011}$. Prove $x_1 = x_2 = \dots = x_{2011}$.