IMO 1988 SL 17

Origin: ISR

Problem

In the convex pentagon ABCDE, the sides BC, CD, DE have the same length. Moreover, each diagonal of the pentagon is parallel to a side (AC is parallel to DE, BD is parallel to AE, etc.). Prove that ABCDE is a regular pentagon.

Solution

Let AC and AD meet BE in R, S, respectively. Then by the conditions of the problem, \angleAEB = \angleEBD = \angleBDC = \angleDBC = \angleADB = \angleEAD = \alpha, \angleABE = \angleBEC = \angleECD = \angleCED = \angleACE = \angleBAC = \beta, \angleBCA = \angleCAD = \angleADE = \gamma. Since \angleSAE = \angleSEA, it follows that AS = SE, and analogously BR = RA. But BSDC and REDC are parallelograms; hence BS = CD = RE, giving us BR = SE and AR = AS. Then also AC = AD, because RS \parallel CD. We deduce that 2\beta = \angleACD = \angleADC = 2\alpha, i.e., \alpha = \beta. It will be sufficient to show that \alpha = \gamma, since that will imply \alpha = \beta = \gamma = 36◦. We have that the sum of the interior angles of ACD is 4\alpha+\gamma = 180◦. We have sin \gamma sin \alpha = AE DE = AE CD = AE RE = sin(2\alpha + \gamma) sin(\alpha + \gamma) , i.e., cos \alpha−cos(\alpha+2\gamma) = 2 sin \gamma sin(\alpha+\gamma) = 2 sin \alpha sin(2\alpha+\gamma) = cos(\alpha+ \gamma) −cos(3\alpha + \gamma). From 4\alpha + \gamma = 180◦we obtain −cos(3\alpha + \gamma) = cos \alpha. Hence cos(\alpha + \gamma) + cos(\alpha + 2\gamma) = 2 cos \gamma 2 cos 2\alpha + 3\gamma = 0, so that 2\alpha + 3\gamma = 180◦. It follows that \alpha = \gamma. Second solution. We have \angleBEC = \angleECD = \angleDEC = \angleECA = \angleCAB, and hence the trapezoid BAEC is cyclic; consequently, AE = BC. Similarly AB = ED, and ABCD is cyclic as well. Thus ABCDE is cyclic and has all sides equal; i.e., it is regular.