IMO 1994 SL G2

Origin: UKR | Category: Geometry

Problem

ABCD is a quadrilateral with BC parallel to AD. M is the midpoint of CD, P that of MA and Q that of MB. The lines DP and CQ meet at N. Prove that N is not outside triangle ABM.8

Solution

Although it does not seem to have been noticed at the jury, the state- ment of the problem is false. For A(0, 0), B(0, 4), C(1, 4), D(7, 0), we have M(4, 2), P(2, 1), Q(2, 3) and N(9/2, 1/2) ̸\in\triangleABM. The official solution, if it can be called so, actually shows that N lies inside ABCD and goes as follows: The case AD = BC is trivial, so let AD > BC. Let L be the midpoint of AB. Complete the parallelograms ADMX and BCMY . Now N = DX \capCY , so let CY and DX intersect AB at K and H respectively. From LX = LY and HL LX = HA AD < LA AD < KB AD < KB BC = KL LY we get HL < KL, and the statement follows.