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.

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