Saudi Arabian Mathematical Competitions, 2012

The relations are equivalent to
$$a + \frac{2}{3}m_b = b + \frac{2}{3}m_a \quad \text{and} \quad b + \frac{2}{3}m_c = c + \frac{2}{3}m_b.$$
We will prove that if $a \le b$, then $m_a \ge m_b$. Indeed, we have
$$m_a^2 - m_b^2 = \frac{2(b^2 + c^2) - a^2}{4} - \frac{2(a^2 + c^2) - b^2}{4} = \frac{3}{4}(b^2 - a^2) \ge 0, \\ \text{and hence } m_a \ge m_b.$$
It follows that if $a \le b \le c$, then $m_a \ge m_b \ge m_c$. From the given relations we have
$$a - b = \frac{2}{3}(m_a - m_b) \ge 0 \quad \text{and} \quad b - c = \frac{2}{3}(m_b - m_c) \ge 0.$$
That is, $a \ge b \ge c$, and thus $a = b = c$.

We shall use the notation in the following figure.

The relations in the problem imply that the triangles $AGB$, $BGC$, $CGA$ have the same perimeter, and hence the same semiperimeter $s'$. Also, these triangles have the same areas. From Heron's formula it follows that
$$s'(s' - a)(s' - b')(s' - c') = s'(s' - a')(s' - b)(s' - c'),$$
so $ab' = a'b$. Similarly, $ac' = a'c$ and $bc' = b'c$. Using these relations and the equality of the areas of $AGB$, $BGC$, $CGA$, we get $\widehat{BAG} = \widehat{BCG}$, $\widehat{CBG} = \widehat{CAG}$, and $\widehat{ABG} = \widehat{ACG}$. Denote these angles by $\alpha$, $\beta$, $\gamma$ respectively.
We have $2(\alpha + \beta + \gamma) = 180^\circ$, so $\alpha + \beta + \gamma = 90^\circ$. It follows that $\widehat{BB'C} = 90^\circ$, that is $BB'$ is altitude. Similarly, the other medians are altitudes, hence triangle $ABC$ is equilateral.