IMO 2003 SL G3

Origin: IND | Category: Geometry

Problem

Let ABC be a triangle and let P be a point in its interior. Denote by D, E, F the feet of the perpendiculars from P to the lines BC, CA, and AB, respectively. Suppose that AP 2 + PD2 = BP 2 + PE2 = CP 2 + PF 2. Denote by IA, IB, IC the excenters of the triangle ABC. Prove that P is the circumcenter of the triangle IAIBIC.

Solution

From the given equality we see that 0 = (BP 2 + PE2) −(CP 2 + PF 2) = BF 2 −CE2, so BF = CE = x for some x. Similarly, there are y and z such that CD = AF = y and BD = AE = z. It is easy to verify that D, E, and F must lie on the segments BC, CA, AB. Denote by a, b, c the length of the segments BC, CA, AB. It follows that a = z + y, b = z + x, c = x + y, so D, E, F are the points where the excircles touch the sides of \triangleABC. Hence P, D, and IA are collinear and \anglePIAC = \angleDIAC = 90◦−180◦−\angleACB = \angleACB . In the same way we obtain that \anglePIBC = \angleACB and PIB = PIA. Analogously, we get PIC = PIB, which implies that P is the circumcenter of the triangle IAIBIC.