Romanian Mathematical Olympiad

Recall that $m_a^2=\frac{1}{4}(2(b^2+c^2)-a^2)$, together with the other similar formulas.

Assume that the squares of the lengths of the sides are in arithmetic progression, for example $2b^2=a^2+c^2$. Then $m_a^2=\frac{3}{4}{c}^2$, $m_b^2=\frac{3}{4}{b}^2$, $m_c^2=\frac{3}{4}{a}^2$, implying that $$\frac{m_a}{c}=\frac{m_b}{b}=\frac{m_c}{a}=\frac{\sqrt{3}}{2}.$$ Obviously, the triangles are similar.

Conversely, assume the triangles are similar. Let $a\le b\le c$. Then $m_a\ge m_b\ge m_c$, so $$\frac{m_a}{c}=\frac{m_b}{b}=\frac{m_c}{a}=k.$$ From the equality $m_a^2+m_b^2+m_c^2=\frac{3}{4}(a^2+b^2+c^2)$ we get $k=\frac{\sqrt{3}}{2}$. Hence $4m_b^2=3b^2$, which yields $2b^2=a^2+c^2$.