THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Using the usual notations for a triangle, we have: $$\begin{array}{rl}\overline{GI}& =\overline{AI}-\overline{AG}=\left(\frac{b}{a+b+c}-\frac{1}{3}\right)\overline{AB}-\left(\frac{c}{a+b+c}-\frac{1}{3}\right)\overline{AC}\\ & =\frac{1}{3(a+b+c)}\left((2b-a-c)\overline{AB}+(2c-a-b)\overline{AC}\right).\end{array}$$ Then $GI\perp BC⟺\overline{GI}\cdot \overline{BC}=0$ is equivalent with $$((2b-a-c)\overline{AB}+(2c-a-b)\overline{AC})\cdot (\overline{AC}-\overline{AB})=0.$$ Because $\overline{AB}\cdot \overline{AB}=c^2$, $\overline{AC}\cdot \overline{AC}=b^2$ and $2\overline{AB}\cdot \overline{AC}=b^2+c^2-a^2$, the above equality is equivalent with $3(b-c)(b^2+c^2-a^2)-2(2b-a-c)c^2+2(2c-a-b)b^2=0$ or $(b-c)(a+b+c)(-3a+b+c)=0$. But $b\ne c$ and $a+b+c>0$, and thus $GI\perp BC⟺b+c=3a$.