Let AP, BP, CP be the cevians of a point P in the plane of a triangle ABC. 

 Let A<sub>B</sub> be a point on BP and A<sub>C</sub> a point on CP such that the triangle AA<sub>B</sub>A<sub>C</sub> is equilateral. Define B<sub>C</sub>, B<sub>A</sub>, C<sub>A</sub> and C<sub>B</sub>  cyclically. 

X(5618) is the only point on the circumcircle such that the lines  A<sub>B</sub>A<sub>C</sub>, B<sub>C</sub>B<sub>A</sub>, C<sub>B</sub>C<sub>A</sub> are concurrent. 

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 Moreover, the centers of the three equilateral triangles are collinear with P; denote their line by L(P).  If P is on the circumcircle of ABC, then L(P) passes through X(110).  &nbsp;&nbsp; (Angel Montesdeoca, November 3, 2013)