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X(2), X(4), X(69), X(74), X(94), X(146), X(323), X(1138), X(1272), X(2071), X(3260) vertices of the antimedial triangle P (pivot) = tX74 = isotomic conjugate of X(74) = X(3260) vertices of the cevian triangle of P intersections of the circumcircle and the de Longchamps axis 

K279 is the isogonal transform of K495 = pK(X32, X30) and the anticomplement of K489. K279 meets the circumcircle at the same points as pK(X6, X323). 

1. Let M be a point. The parallel at M to the line AX(74) meets BC at Ma. Define Mb, Mc similarly. ABC and MaMbMc are perspective if and only if P lies on K255. The locus of the perspector is K279. See also Central cubics for a generalization. 2. Let A' B' C' be the cevian triangle of a point P. The circumcircles (Oa), (Ob), (Oc) of AB'C', BC'A', CA'B' are concurrent at M. Denote by O* the circumcenter of OaObOc. K279 is the locus of point P such that O* lies on the Euler line (Angel Montesdeoca, private message 20130806, see also Anopolis #758). Remark : The Euler line can be replaced by any line through O and one always gets an isotomic pK with pivot on the de Longchamps axis. Any other line will give a circumcubic and, in particular, a circular cubic with the perpendicular bisector of OH. 
