IMO 2004 SL G8

Origin: SMN | Category: Geometry

Problem

A cyclic quadrilateral ABCD is given. The lines AD and BC intersect at E, with C between B and E; the diagonals AC and BD intersect at F. Let M be the midpoint of the side CD, and let N ̸= M be a point on the circumcircle of the triangle ABM such that AN/BN = AM/BM. Prove that the points E, F, and N are collinear.

Solution

To start with, note that point N is uniquely determined by the imposed properties. Indeed, f(X) = AX/BX is a monotone function on both arcs AB of the circumcircle of \triangleABM. Denote by P and Q respectively the second points of intersection of the line EF with the circumcircles of \triangleABE and \triangleABF. The prob- lem is equivalent to showing that N \inPQ. In fact, we shall prove that N coincides with the midpoint N of segment PQ. The cyclic quadrilaterals APBE, AQBF, and ABCD yield \angleAPQ = 180◦−\angleAPE = 180◦−\angleABE = \angleADC and \angleAQP = \angleAQF = \angleABF = \angleACD. It follows that \triangleAPQ ∼ \triangleADC, and conse- quently \triangleANP ∼\triangleAMD. Analo- A B C D E F M P Q N Ω gously \triangleBNP ∼\triangleBMC. Therefore AN/AM = PQ/DC = BN/BM, i.e., AN/BN = AM/BM. Moreover, \angleANB = \angleANP + \anglePNB = \angleAMD + \angleBMC = 180◦−\angleAMB, which means that point N lies on

the circumcircle of \triangleAMB. By the uniqueness of N, we conclude that N \equivN, which completes the solution.