IMO 1999 SL G7

Origin: ARM | Category: Geometry

Problem

The point M inside the convex quadrilateral ABCD is such that MA = MC, \angleAMB = \angleMAD+\angleMCD, \angleCMD = \angleMCB+\angleMAB. Prove that AB \cdot CM = BC \cdot MD and BM \cdot AD = MA \cdot CD.

Solution

Let us construct a convex quadrilateral PQRS and an interior point T such that \trianglePTQ ∼= \triangleAMB, \triangleQTR ∼\triangleAMD, and \trianglePTS ∼\triangleCMD. We then have TS = MD\cdotPT MC = MD and T R T S = T R\cdotT Q\cdotT P T Q\cdotT P\cdotT S = MD\cdotMB\cdotMC MA\cdotMA\cdotMD = MB MC (using MA = MC). We also have \angleSTR = \angleBMC and therefore \triangleRTS ∼\triangleBMC. Now the relations between angles become \angleTPS + \angleTQR = \anglePTQ and \angleTPQ + \angleTSR = \anglePTS, implying that PQ \parallelRS and QR \parallelPS. Hence PQRS is a parallelogram and hence AB = PQ = RS and QR = PS. It follows that BC MC = RS T S = AB MD ⇒AB \cdot CM = BC \cdot MD and AD\cdotBM AM

AD\cdotQT AM = QR = PS = CD\cdotT S MD = CD ⇒BM \cdot AD = MA \cdot CD.