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Triangle sides divided in 3 parts and concurrent lines

Source: Macedonian TST for IMO 2013 - P1 day 1

March 28, 2021
geometrycircumcircle

Problem Statement

The points A1,A2,B1,B2,C1,C2A_{1},A_{2},B_{1},B_{2},C_{1},C_{2} are on the sides ABAB, BCBC and ACAC of an acute triangle ABCABC such that AA1=A1A2=A2B=13ABAA_{1} = A_{1}A_{2} = A_{2}B = \frac{1}{3} AB, BB1=B1B2=B2C=13BCBB_{1} = B_{1}B_{2} = B_{2}C = \frac{1}{3}BC and CC1=C1C2=C2A=13ACCC_{1} = C_{1}C_{2} = C_{2}A = \frac{1}{3} AC. Let kA,kBk_{A}, k_{B} and kCk_{C} be the circumcircles of the triangles AA1C2AA_{1}C_{2}, BB1A2BB_{1}A_{2} and CC1B2CC_{1}B_{2} respectively. Furthermore, let aBa_{B} and aCa_{C} be the tangents to kAk_{A} at A1A_{1} and C2C_{2}, bCb_{C} and bAb_{A} the tangents to kBk_{B} at B1B_{1} and A2A_{2} and cAc_{A} and cBc_{B} the tangents to kCk_{C} at C1C_{1} and B2B_{2}. Show that the perpendicular lines from the intersection points of aBa_{B} and bAb_{A}, bCb_{C} and cBc_{B}, cAc_{A} and aCa_{C} to ABAB, BCBC and CACA respectively are concurrent.