67th Romanian Mathematical Olympiad

As usual, we could suppose that $|a|=|b|=|c|=1$. The equality in the hypothesis is equivalent with $(a+b+c{)}^2=ab+bc+ca$, and thus $(a+b+c{)}^2=abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$, which is the same with $(a+b+c{)}^2=abc(a+b+c)$. It is clear now that $|a+b+c|\in \{0,1\}$.

If $|a+b+c|=0$, then the orthocenter of the triangle with the vertices whose complex coordinates are $a$, $b$, $c$, coincides with its circumcenter, and thus the triangle is equilateral.

If $|a+b+c|=1$, then $(a+b+c)\overline{(a+b+c)}=1$, and thus $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1,$$ that is $(a+b)(b+c)(c+a)=0$. Then, at least one of the above parenthesis is zero, and thus the triangle has two antipodal vertices, that is it is a right angled triangle.