Romanian Mathematical Olympiad

The given similarity rewrites as $$\frac{m-b}{a-b}=\frac{n-c}{b-c}=\frac{p-a}{c-a}=k,$$ hence $$m=ka+(1-k)b,$$ $$n=kb+(1-k)c,$$ $$p=kc+(1-k)a.$$ Since the triangle $MNP$ is equilateral, we have $$m+\epsilon n+{\epsilon }^{2}p=0,$$ where $\epsilon =\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}$. Substituting, we infer that $$\begin{array}{rl}0& =k(a+b\epsilon +c{\epsilon }^2)+(1-k)(b+c\epsilon +a{\epsilon }^2)\\ & =k(a+b\epsilon +c{\epsilon }^2)+\frac{1-k}{\epsilon }(a+b\epsilon +c{\epsilon }^2)\\ & =(a+b\epsilon +c{\epsilon }^2)\left(k+\frac{1-k}{\epsilon }\right).\end{array}$$ The triangle $ABC$ is not equilateral, so $a+b\epsilon +c{\epsilon }^2\ne 0$, and consequently $$k=\frac{1}{1-\epsilon }.$$ The equality $m=ka+(1-k)b$ yields $m-a=\epsilon (m-b)$, showing that triangle $AMB$ is isosceles, with an angle $\frac{2\pi }{3}$ and two angles $\frac{\pi }{6}$.