Estonian Mathematical Olympiad

Fig. 40

Denote $AC=b$, $AB=c$, $\angle BAC=\alpha$, and $AE=AP=u$. As $BCEN$ is a parallelogram, $BN=CE=b-u$ and $BN\parallel CE$. The latter implies $\angle NBP=\alpha$ (Fig. 40). As $\angle QAE=\angle QBN$ and $\angle AQE=\angle BQN$, triangles $AQE$ and $BQN$ are similar.

a) The triangles $AEP$ and $BNP$ have areas $\frac{1}{2}{u}^2\sin \alpha$ and $\frac{1}{2}(b-u)(c-u)\sin \alpha$, respectively. Thus $u^2=(b-u)(c-u)$, whence $\frac{c-u}{u}=\frac{u}{b-u}$. Similarity of triangles $AQE$ and $BQN$ implies $\frac{AQ}{BQ}=\frac{AE}{BN}$ which, after defining $BQ=x$, rewrites to $\frac{c-x}{x}=\frac{u}{b-u}$. This equality reduces to a linear equation of $x$, meaning that it has only one root. By equality $\frac{c-u}{u}=\frac{u}{b-u}$, $x=u$ must be the only root. Thus $BQ=u=AP$, whence the midpoint of the side $AB$ coincides with the midpoint of the line segment $PQ$.

Fig. 41

b) Let $F$ be the midpoint of the side $AB$ and let $D$ be the point of intersection of the bisector of angle $A$ with side $BC$ (Fig. 41). By angle bisector theorem, $\frac{BD}{CD}=\frac{AB}{AC}$. Since $NE$ and $BC$ are parallel, triangles $ABC$ and $AQE$ are similar, whence also triangles $ABC$ and $BQN$ are similar. This and part a) of the problem together imply $\frac{CE}{EA}=\frac{BN}{BQ}=\frac{AC}{AB}$. Consequently, $\frac{AF}{FB}\cdot \frac{BD}{DC}\cdot \frac{CE}{EA}=1\cdot \frac{AB}{AC}\cdot \frac{AC}{AB}=1$, giving the desired result by Ceva's theorem.