IMO 2000 SL G2

Origin: RUS | Category: Geometry

Problem

Two circles G1 and G2 intersect at M and N. Let AB be the line tangent to these circles at A and B, respectively, such that M lies closer to AB than N. Let CD be the line parallel to AB and passing through M, with C on G1 and D on G2. Lines AC and BD meet at E; lines AN and CD meet at P; lines BN and CD meet at Q. Show that EP = EQ.

Solution

Let K be the intersection point of the lines MN and AB. Since KA2 = KM \cdot KN = KB2, it follows that K is the midpoint of the segment AB, and consequently M is the midpoint of AB. Thus it will be enough to show that EM \perp PQ, or equivalently that EM \perp AB. However, since AB is tangent to the circle G1 we have \angleBAM = \angleACM = \angleEAB, and similarly M N B A C D K E P Q \angleABM = \angleEBA. This implies that the triangles EAB and MAB are congruent. Hence E and M are symmetric with respect to AB; hence EM \perpAB. Remark. The proposer has suggested an alternative version of the prob- lem: to prove that EN bisects the angle CND. This can be proved by noting that EANB is cyclic.