IMO 1998 SL 1

Origin: LUX

Problem

A convex quadrilateral ABCD has perpendicular diagonals. The perpendicular bisectors of AB and CD meet at a unique point P inside ABCD. Prove that ABCD is cyclic if and only if triangles ABP and CDP have equal areas.

Solution

We begin with the following observation: Suppose that P lies in \triangleAEB, where E is the intersection of AC and BD (the other cases are similar). Let M, N be the feet of the perpendiculars from P to AC and BD respectively. We have SABP = SABE −SAEP −SBEP = 1 2(AE \cdot BE −AE \cdot EN −BE \cdot EM) = 1 2(AM \cdotBN −EM \cdotEN). Similarly, SCDP = 1 2(CM \cdotDN −EM \cdot EN). Therefore, we obtain SABP −SCDP = AM \cdot BN −CM \cdot DN . (1) Now suppose that ABCD is cyclic. Then P is the circumcenter of ABCD; hence M and N are the midpoints of AC and BD. Hence AM = CM and BN = DN; thus (1) gives us SABP = SCDP . On the other hand, suppose that ABCD is not cyclic and let w.l.o.g. A B C D E P M N PA = PB > PC = PD. Then we must have AM > CM and BN > DN, and consequently by (1), SABP > SCDP . This proves the other implication. Second solution. Let F and G denote the midpoints of AB and CD, and assume that P is on the same side of FG as B and C. Since PF \perpAB, PG \perpCD, and \angleFEB = \angleABE, \angleGEC = \angleDCE, a direct computa- tion yields \angleFPG = \angleFEG = 90◦+ \angleABE + \angleDCE. Taking into account that SABP = 1 2AB \cdot FP = FE \cdot FP, we note that SABP = SCDP is equivalent to FE \cdot FP = GE \cdot GP, i.e., to FE/EG = GP/PF. But this last is equivalent to triangles EFG and PGF being similar, which holds if and only if EFPG is a parallelogram. This last is equivalent to \angleEFP = \angleEGP, or 2\angleABE = 2\angleDCE. Thus SABP = SCDP is equivalent to ABCD being cyclic. Remark. The problems also allows an analytic solution, for example putting the x and y axes along the diagonals AC and BD.