Romanian Mathematical Olympiad - District Round

a) If $I_A$ is the point of intersection of the exterior bisectors of the angles $\angle B$ and $\angle C$, then $I_A$ is inside the angle $\angle A$ and is equidistant from the sides $AB$ and $BC$, respectively of $BC$ and $AC$. Through transitivity, $I_A$ is equidistant from the sides $AB$ and $AC$ of the angle $\angle A$, is inside the angle $\angle A$, so it is on the interior bisector of the angle $\angle A$.

b) From the condition $I_A M + I_A P = I_A N$ we deduce that the quadrilateral $I_A M N P$ is a parallelogram. Also from point a) we have $I_A M = I_A N = I_A P$, so $I_A M N P$ is the rhombus and the triangles $I_A N P$ and $I_A M N$ are equilateral.
Thus, the inscribed quadrilateral $API_A M$ it has $\angle P I_A M = 120^\circ$, so $\angle A = 60^\circ$,
On the other hand, $C$ is exterior angle of the inscribed quadrilateral $I_A M C N$ so $\angle C = \angle M I_A N = 60^\circ$, and $B$ is exterior angle of the inscribed quadrilateral $I_A P B N$ so $\angle B = \angle P I_A N = 60^\circ$. Thus, triangle $ABC$ has all angles of $60^\circ$, so it is equilateral.