Croatian Mathematical Olympiad

Question

A point T is chosen inside the triangle ABC. Let A_1, B_1 and C_1 be the reflections of T across the lines BC, CA and AB, respectively. The lines A_1T, B_1T and C_1T intersect the circle k circumscribed to the triangle A_1B_1C_1 again at A_2, B_2 and C_2, respectively. Prove that the lines AA_2, BB_2 and CC_2 are concurrent on k. (IMO Shortlist 2018) FIGURE_0

Accepted Answer

Let K be the intersection of CC_2 and k. FIGURE_0 Since CB and CA are the bisectors of TA_1 and TB_1, respectively, the point C is the circumcentre of the triangle A_1TB_1. Hence, (CA_1, CB) = (CB, CT) = (B_1A_1, B_1T) = (B_1A_1, B_1B_2). Observing the circle k, we have (B_1A_1, B_1B_2) = (C_2A_1, C_2B_2) and (CA_1, CB) = (B_1A_1, B_1B_2) = (C_2A_1, C_2B_2). Similarly, we get (BA_1, BC) = (B_2A_1, B_2C_2), hence the triangles A_1BC and A_1B_2C_2 are similar, so as the triangles A_1BB_2 and A_1CC_2, from which it follows that (C_2C, C_2A_1) = (B_2B, B_2A_1) and the point K lies on BB_2. Analogously, we show that the point K lies on AA_2, and the proof is finished.