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locus of circumcenters 2<B_1A_1C_1+<BAC = 180^o(VII Soros Olympiad R1 9.8)

Source:

July 4, 2021
geometryanglesLocusCircumcenter

Problem Statement

Given a triangle ABCABC. On its sides BCBC , CACA and ABAB , the points A1A_1 , B1B_1 and C1C_1 are taken, respectively , such that 2B1A1C1+BAC=180o2 \angle B_1 A_1 C_1 + \angle BAC = 180^o , 2A1C1B1+ACB=180o2 \angle A_1 C_1 B_1 + \angle ACB = 180^o , 2C1B1A1+CBA=180o2 \angle C_1 B_1 A_1 + \angle CBA = 180^o . Find the locus of the centers of the circles circumscribed about the triangles A1B1C1A_1 B_1 C_1 (all possible such triangles are considered).