Saudi Arabia Mathematical Competitions 2012

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

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Alternative solution.

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 $s^{\prime }$. Also, these triangles have the same areas. From Heron's formula it follows that $$s^{\prime }(s^{\prime }-a)(s^{\prime }-b^{\prime })(s^{\prime }-c^{\prime })=s^{\prime }(s^{\prime }-a^{\prime })(s^{\prime }-b)(s^{\prime }-c^{\prime }),$$ so $a{b}^{\prime }=a^{\prime }b$. Similarly, $a{c}^{\prime }=a^{\prime }c$ and $b{c}^{\prime }=b^{\prime }c$. 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 $\widehat{B{B}^{\prime }C}=90^{\circ }$, that is $B{B}^{\prime }$ is altitude. Similarly, the other medians are altitudes, hence triangle $ABC$ is equilateral.