MathDB
triangle geometry [AP^2 + PD^2 = ...]

Source: australian tst 1, IMO ShortList 2003, geometry problem 3; Indian IMOTC 2004 Day 2 Problem 1

May 12, 2004
geometrycircumcircleCircumcenterTriangleIMO Shortlistexcentersgeometry solved

Problem Statement

Let ABCABC be a triangle and let PP be a point in its interior. Denote by DD, EE, FF the feet of the perpendiculars from PP to the lines BCBC, CACA, ABAB, respectively. Suppose that AP2+PD2=BP2+PE2=CP2+PF2.AP^2 + PD^2 = BP^2 + PE^2 = CP^2 + PF^2. Denote by IAI_A, IBI_B, ICI_C the excenters of the triangle ABCABC. Prove that PP is the circumcenter of the triangle IAIBICI_AI_BI_C.
Proposed by C.R. Pranesachar, India
Back to ProblemsView on AoPS