# IMO 1998 Problem 1

Problem 1 of the International Mathematical Olympiad 1998

In the convex quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel. Suppose that the point P, where the perpendicular bisectors of AB and DC meet, is inside ABCD. Prove that ABCD is a cyclic quadrilateral if and only if the triangles ABP and CDP have equal areas.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to my children Alan Huy and Cathy Diem) Let AC intercept BD at J, AB intercept CD at I, E and F be the midpoints of AB and CD, respectively. We have

∠EAJ = ∠EJA, ∠EBJ = ∠EJB, ∠FDJ = ∠FJD, ∠FJC = ∠FCJ and JE = ½ AB, JF = ½ CD. Triangles ABP and CDP having equal areas give us

```		PE × AB = PF × CD		or
PE/PF = CD/AB = JF/JE							(*)
```

But ∠EJA + ∠EJB + ∠FJD + ∠FJC = 180° Or ∠EJB + ∠FJC = 180° - ∠EJA - ∠FJD

Therefore, ∠EJF = ∠EJB + 90° + ∠FJC = ∠EBJ + 90° + ∠FCJ = 180° - ∠EIF = ∠EPF

Combined with (*) and the fact that they share segment EF, the triangles JEF and PFE are congruent, and EPFJ is a parallelogram. It follows that PE = JF = DF and PF = JE = AE and the two triangles AEP and PFD are congruent which causes PA = PD or P is the center of the circumcircle passing through A, B, C and D. ABCD is then a cyclic quadrilateral.

Conversely, if ABCD is a cyclic quadrilateral, P is the center of the circumcircle. Since AC is perpendicular to BD, the sum of the angles subtending arcs AB plus CD equal 90°. Therefore, ∠APB + ∠CPD = 180° or ∠APE + ∠FPD = 90° or ∠APE = ∠DPF and the two triangles APE and DPF are congruent (similar triangles with PA = PD). Therefore, triangles ABP and CDP with each having twice the areas of the triangles APE and DPF, respectively, have equal areas.