IMO 2012 Shortlist G1

Category: Geometry

Problem

In the triangle ABC the point J is the center of the excircle opposite to A. This excircle is tangent to the side BC at M, and to the lines AB and AC at K and L respectively. The lines LM and BJ meet at F, and the lines KM and CJ meet at G. Let S be the point of intersection of the lines AF and BC, and let T be the point of intersection of the lines AG and BC. Prove that M is the midpoint of ST. Solution. Let α = ∠CAB, β = ∠ABC and γ = ∠BCA. The line AJ is the bisector of ∠CAB, so ∠JAK = ∠JAL = α . By ∠AKJ = ∠ALJ = 90◦ the points K and L lie on the circle ω with diameter AJ. The triangle KBM is isosceles as BK and BM are tangents to the excircle. Since BJ is the bisector of ∠KBM, we have ∠MBJ = 90◦ − β and ∠BMK = β . Likewise ∠MCJ = 90◦ − γ and ∠CML = γ . Also ∠BMF = ∠CML, therefore ∠LFJ = ∠MBJ − ∠BMF =  90◦ − β  − γ

α = ∠LAJ. Hence F lies on the circle ω. (By the angle computation, F and A are on the same side of BC.) Analogously, G also lies on ω. Since AJ is a diameter of ω, we obtain ∠AFJ = ∠AGJ = 90◦ . A B C G F S T K M L ω J β γ α α α α The lines AB and BC are symmetric with respect to the external bisector BF. Because AF ⊥ BF and KM ⊥ BF, the segments SM and AK are symmetric with respect to BF, hence SM = AK. By symmetry TM = AL. Since AK and AL are equal as tangents to the excircle, it follows that SM = TM, and the proof is complete. Comment. After discovering the circle AFKJLG, there are many other ways to complete the solu- tion. For instance, from the cyclic quadrilaterals JMFS and JMGT one can find ∠TSJ = ∠STJ = α 2 . Another possibility is to use the fact that the lines AS and GM are parallel (both are perpendicular to the external angle bisector BJ), so MS MT = AG GT = 1.30