Croatian Mathematical Society Competitions

The triangle $ABC$ is acute-angled, so the points $D$, $E$ and $F$ lie on the sides $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$, respectively. Since $|AB|>|AC|$, we have $|BD|>|CD|$.

Furthermore, from $\angle ACB<\angle ABC$ it follows that $C$ lies between $B$ and $P$. Also, $C$ lies between $A$ and $Q$, while $R$ lies on the segment $\overline{BF}$.



Let $H$ be the orthocentre of the triangle $ABC$, and let $M$ be the midpoint of its side $\overline{BC}$. The lines $DR$ and $EF$ are parallel, hence $\angle DRF=\angle EFA$. The quadrilateral $AFHE$ is cyclic ($\angle AFH=\angle AEH=90^{\circ }$), so we have $\angle EFA=\angle EHA$. The quadrilateral $BDHF$ is cyclic as well ($\angle BDH=\angle BFH=90^{\circ }$), hence $\angle EHA=\angle DHB=\angle DFB=\angle DFR$. Now from $\angle DRF=\angle DFR$ it follows that $|DR|=|DF|$.

Similarly, the lines $DQ$ and $FE$ are parallel, hence $\angle DQE=\angle FEA$. The quadrilateral $AFHE$ is cyclic, so we have $\angle FEA=\angle FHA$. The quadrilateral $CEHD$ is cyclic as well ($\angle CEH=\angle CDH=90^{\circ }$), hence $\angle FHA=\angle DHC=\angle DEC=\angle DEQ$. Now from $\angle DQE=\angle DEQ$ it follows that $|DQ|=|DE|$.

The points $M$, $F$, $E$ and $D$ lie on the same circle (called the nine-point circle or the Feuerbach's circle), hence $\angle FMD=180^{\circ }-\angle FED=\angle PED$. Also, $\angle FDH=\angle FBH=\angle ABE=90^{\circ }-\angle CAB=\angle ACF=\angle ECH=\angle EDH$.

Therefore, $\angle FDM=90^{\circ }-\angle FDH=90^{\circ }-\angle EDH=\angle PDE$, so the triangles $FDM$ and $PDE$ are similar. Now we conclude that $$\frac{|DF|}{|DM|}=\frac{|DP|}{|DE|},$$ from which it follows that $|DF|\cdot |DE|=|DP|\cdot |DM|$, i.e. $$|DR|\cdot |DQ|=|DP|\cdot |DM|,$$ meaning that the quadrilateral $PQMR$ is cyclic.

Let $k$ be its circumscribed circle. Since $\angle NQP+\angle NRP<180^{\circ }$, $N$ lies inside the circle $k$, i.e. on the segment $\overline{CM}$, and $N\ne M$ holds. Since $M$ is the midpoint of the segment $\overline{BC}$, we finally get $|BN|>|CN|$.