Let AP, BP, CP be the cevians of a point P in the plane of a triangle ABC. Let AB be a point on BP and AC a point on CP such that the triangle AABAC is equilateral. Define BC, BA, CA and CB cyclically. X(5618) is the only point on the circumcircle such that the lines ABAC, BCBA, CBCA are concurrent.

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).    (Angel Montesdeoca, November 3, 2013)