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3

Part of 2013 IMAC Arhimede

Problems(1)

romanian concyclic, starting with <B=120^o, angle bisectors and 3 incenters

Source: IMAC Arhimede 2013 p3

9/26/2018
Let ABCABC be a triangle with ∠ABC=120o\angle ABC=120^o and triangle bisectors (AA1),(BB1),(CC1)(AA_1),(BB_1),(CC_1), respectively. B1F⊥A1C1B_1F \perp A_1C_1, where F∈(A1C1)F\in (A_1C_1). Let R,IR,I and SS be the centers of the circles which are inscribed in triangles C1B1F,C1B1A1,A1B1FC_1B_1F,C_1B_1A_1, A_1B_1F, and B1S∩A1C1={Q}B_1S\cap A_1C_1=\{Q\}. Show that R,I,S,QR,I,S,Q are on the same circle.
geometryincenterangle bisectorConcyclic