IMO 2004 SL G2

Origin: KAZ | Category: Geometry

Problem

The circle \Gamma and the line ℓdo not intersect. Let AB be the diameter of \Gamma perpendicular to ℓ, with B closer to ℓthan A. An arbitrary point C ̸= A, B is chosen on \Gamma. The line AC intersects ℓat D. The line DE is tangent to \Gamma at E, with B and E on the same side of AC. Let BE intersect ℓat F, and let AF intersect \Gamma at G ̸= A. Prove that the reflection of G in AB lies on the line CF.

Solution

Let H be the reflection of G about AB (GH \parallel ℓ). Let M be the intersection of AB and ℓ. Since \angleFEA = \angleFMA = 90◦, it follows that AEMF is cyclic and hence \angleDFE = \angleBAE = \angleDEF. The last equality holds because DE is tangent to \Gamma. It follows that DE = DF and hence DF 2 = DE2 = DC \cdot DA (the power of D with re- A B M E F G D C H \Gamma ℓ spect to \Gamma). It then follows that \angleDCF = \angleDFA = \angleHGA = \angleHCA. Thus it follows that H lies on CF as desired.