IMO 2003 SL G4

Origin: ARM | Category: Geometry

Problem

Let \Gamma1, \Gamma2, \Gamma3, \Gamma4 be distinct circles such that \Gamma1, \Gamma3 are externally tangent at P, and \Gamma2, \Gamma4 are externally tangent at the same point P. Suppose that \Gamma1 and \Gamma2; \Gamma2 and \Gamma3; \Gamma3 and \Gamma4; \Gamma4 and \Gamma1 meet at A, B, C, D, respectively, and that all these points are different from P. Prove that AB \cdot BC AD \cdot DC = PB2 PD2 .

Solution

Apply an inversion with center at P and radius r; let < X denote the image of X. The circles \Gamma1, \Gamma2, \Gamma3, \Gamma4 are transformed into lines @ \Gamma1, @ \Gamma2, @ \Gamma3, @ \Gamma4, where @ \Gamma1 \parallel@ \Gamma3 and @ \Gamma2 \parallel@ \Gamma4, and therefore <A <B <C <D is a parallelogram. Further, we have AB = r2 P < A\cdotP <B <A <B, BC = r2 P <B\cdotP <C <B <C, CD = r2 P <C\cdotP <D <C <D, DA = r2 P <D\cdotP < A <D <A and PB = r2 P <B , PD = r2 P <D. The equality to be proven becomes P <D2 P <B2 \cdot <A <B \cdot <B <C <A <D \cdot <D <C = P <D2 P <B2 , which holds because <A <B = <C <D and <B <C = <D <A.