74th Romanian Mathematical Olympiad

On the other hand, from $\angle APB=120^{\circ }$ and $\angle APF=30^{\circ }$, it follows that $\angle FPB=90^{\circ }$, hence the triangle $PBF$ is a $30^{\circ }-60^{\circ }-90^{\circ }$ triangle, from which $BF=2FP$, (2). Using (1) and (2) we conclude $BF=2FA$, (3). Denote $B{B}^{\prime }$ the median from $B$ of the triangle $ABC$. Since $G$ is the centroid of the triangle $ABC$, we have $BG=2G{B}^{\prime }$ (4). The relations (3) and (4) lead, according to the converse of Thales' theorem, to $FG\parallel AC$, thus points $P,F$ and $G$ are collinear. We obtain $\angle GPN=30^{\circ }$. Similarly, it follows that $\angle GNP=30^{\circ }$ hence the triangle $GNP$ is isosceles, with $GP=GN$. Since $AFGE$ is a parallelogram, $GP=GF+FP=AE+FP=\frac{AB+AC}{3}$, from which the conclusion follows.