Saudi Arabia Mathematical Competitions

Let $A^{\prime }$, $B^{\prime }$, $C^{\prime }$ be the midpoints of sides $BC$, $CA$, $AB$, respectively. Construct the parallelogram $A{C}^{\prime }C{C}^{\prime \prime }$. The points $C^{\prime }$, $B^{\prime }$, $C^{\prime \prime }$ are collinear, hence $B{A}^{\prime }{C}^{\prime \prime }{B}^{\prime }$ is also a parallelogram, that is $A^{\prime }{C}^{\prime \prime }=B{B}^{\prime }$.



The desired triangle is $A{A}^{\prime }{C}^{\prime \prime }$, and $A{A}^{\prime }=m_a$, $A^{\prime }{C}^{\prime \prime }=m_b$, $A{C}^{\prime \prime }=m_c$.



b. Recall the median formula $$m_a^2=\frac{1}{4}\left(2(b^2+c^2)-a^2\right)$$ together with the other two similar formulas for $m_b$ and $m_c$. Assume that the squares 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 $2b^2=a^2+c^2$.