IMO 1999 SL G5

Origin: FRA | Category: Geometry

Problem

Let ABC be a triangle, Ωits incircle and Ωa, Ωb, Ωc three circles three circles orthogonal to Ωpassing through B and C, A and C, and A and B respectively. The circles Ωa, Ωb meet again in C′; in the same way we obtain the points B′ and A′. Prove that the radius of the circumcircle of A′B′C′ is half the radius of Ω.

Solution

Let Ω(I, r) be the incircle of \triangleABC. Let D, E, and F denote the points where Ωtouches BC, AC, and AB, respectively. Let P, Q, and R denote the midpoints of EF, DF, and DE respectively. We prove that Ωa passes through Q and R. Since \triangleIQD ∼\triangleIDB and \triangleIRD ∼\triangleIDC, we obtain IQ \cdot IB = IR \cdot IC = r2. We conclude that B, C, Q, and R lie on a single circle \Gammaa. Moreover, since the power of I with respect to \Gammaa is r2, it follows for a tangent IX from I to \Gammaa that X lies on Ωand hence Ωis perpendicular to \Gammaa. From the uniqueness of Ωa it follows that Ωa = \Gammaa. Thus Ωa contains Q and R. Similarly Ωb contains P and R and Ωc contains P and Q. Hence, A′ = P, B′ = Q and C′ = R. Therefore the radius of the circumcircle of \triangleA′B′C′ is half the radius of Ω.