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Bulgaria geometry
Problem
The points , , are chosen on the sides , , of a triangle so that and . The lines and meet the line through , parallel to , at , . Let the circumcircles of the triangles and meet at . Given that lies on , show that lies on the incircle of . (Emil Kolev)
Solution
Observe that are concyclic. Let be the incenter of triangle . Then and perpendicularly bisect and respectively. Hence, is the circumcenter of . Let . . Therefore concyclic , . Similarly , so circle is the incircle of . Thus, lies on the incircle of .
Techniques
Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleTangentsAngle chasing