imo.wolfram.com
header_a.gif Wolfram Research  
mainscorescontestantscompetition problemsaboutphotoIMO Facts


41st International Mathematical Olympiad


IMO 2000 Problems and Solutions

Taejon, Korea
July 13-25, 2000



Problem 1

AB is tangent to the circles CAMN and NMBD. M lies between C and D on the line CD, and CD is parallel to AB. The chords NA and CM meet at P; the chords NB and MD meet at Q. The rays CA and DB meet at E. Prove that PE = QE.


Solution

Angle EBA = angle BDM (because CD is parallel to AB) = angle ABM (because AB is tangent at B). So AB bisects EBM. Similarly, BA bisects angle EAM. Hence E is the reflection of M in AB. So EM is perpendicular to AB and hence to CD. So it suffices to show that MP = MQ.

Let the ray NM meet AB at X. XA is a tangent so XA2 = XM.XN. Similarly, XB is a tangent, so XB2 = XM.XN. Hence XA = XB. But AB and PQ are parallel, so MP = MQ.



Back to Competition Problems page