Saudi Arabian IMO Booklet

Solution. To prove that $MP+NQ=BR$, by adding $RC$ to both sides it is equivalent to prove $CQ+MP=BC$. We start by defining point $T$ on the segment $CQ$ such that $BC=CT$, then it is sufficient to prove that $QT=PM$, since triangles $BTC$ and $NRC$ are isosceles. One can get that $BT$ is parallel to $NR$ so $$\angle BTC=\angle RNC=\angle NRC=\angle BPC,$$ thus $BTPC$ is cyclic. Also we have that $$\angle CNA=180^{\circ }-\angle NRC=180^{\circ }-\angle BPC=\angle CPA,$$ this implies that $PNCA$ is cyclic.

$$\angle TPQ=\angle TCB=\angle BQC=\angle TQP,$$ thus $TQ=TP$. Now by angle chasing, $$\angle CPA=180^{\circ }-\angle BPC=180^{\circ }-\angle BTC=180^{\circ }-\angle TBC=\angle TPC,$$ thus we have $\angle CPQ=\angle TPC$ and $\angle PCT=\angle PAM$, which implies that $TPC$ and $MPA$ are congruent, since $PC=AP$. In conclusion, $TP=MP=TQ$, which finishes the proof. □