IMO 1998 SL 2

Origin: POL

Problem

Let ABCD be a cyclic quadrilateral. Let E and F be variable points on the sides AB and CD, respectively, such that AE : EB = CF : FD. Let P be the point on the segment EF such that PE : PF = AB : CD. Prove that the ratio between the areas of triangles APD and BPC does not depend on the choice of E and F.

Solution

If AD and BC are parallel, then ABCD is an isosceles trapezoid with AB = CD, so P is the midpoint of EF. Let M and N be the midpoints of AB and CD. Then MN \parallelBC, and the distance d(E, MN) equals the distance d(F, MN) because B and D are the same distance from MN and EM/BM = FN/DN. It follows that the midpoint P of EF lies on MN, and consequently SAPD : SBPC = AD : BC. If AD and BC are not parallel, then they meet at some point Q. It is plain that \triangleQAB ∼\triangleQCD, and since AE/AB = CF/CD, we also deduce that \triangleQAE ∼\triangleQCF. Therefore \angleAQE = \angleCQF. Further, from these similarities we obtain QE/QF = QA/QC = AB/CD = PE/PF,

which in turn means that QP is the internal bisector of \angleEQF. But since \angleAQE = \angleCQF, this is also the internal bisector of \angleAQB. Hence P is at equal distances from AD and BC, so again SAPD : SBPC = AD : BC. Remark. The part AB \parallelCD could also be regarded as a limiting case of the other part. Second solution. Denote \lambda = AE AB , AB = a, BC = b, CD = c, DA = d, \angleDAB = \alpha, \angleABC = \beta. Since d(P, AD) = c\cdotd(E,AD)+a\cdotd(F,AD) a+c , we have SAPD = cSEAD+aSF AD a+c = \lambdacSABD+(1−\lambda)aSACD a+c . Since SABD = 1 2ad sin \alpha and SACD = 1 2cd sin \beta, we are led to SAPD = acd a+c[\lambda sin \alpha + (1 −\lambda) sin \beta], and analogously SBPC = abc a+c[\lambda sin \alpha + (1 −\lambda) sin \beta]. Thus we obtain SAPD : SBPC = d : b.