17th Mexico 2003 Problem B1 The quadrilateral ABCD has AB parallel to CD. P is on the side AB and Q on the side CD such that AP/PB = DQ/CQ. M is the intersection of AQ and DP, and N is the intersection of PC and QB. Find MN in terms of AB and CD.
MN = AB·CD/(AB+CD) Solution AMP and QMD are similar, so AM/MQ = AP/DQ. PNB and CNQ are similar, so PN/NC = PB/CQ. But AP/DQ = PB/CQ (given), so AM/MQ = PN/NC and hence MN is parallel to AB. Also MN/CD = PM/(PM+MD) = AP/(AP+DQ). Similarly, MN/AB = QM/(QM+MA) = DQ/(AP+DQ). Hence MN/CD + MN/AB = 1. 17th Mexican 2003 © John Scholes jscholes@kalva.demon.co.uk 1 March 2004 Last corrected/updated 1 Mar 04