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∑ a^2 x (y^2  z^2) = 0 or ∑ (c^2 y  b^2 z) y z = 0 

X(2), X(6), X(76), X(194), X(2998) vertices of the antimedial triangle 

Consider a point P = u : v : w and its two Brocardians P1 = 1/w : 1/u : 1/v, P2 = 1/v : 1/w : 1/u. See Tucker cubics for more details. The cevian triangles of P1, P2 are (P1a, P1b, P1c) and (P2a, P2b, P2c). The perpendicular bisectors of the three segments P1xP2x are concurrent (at Q) if and only if P lies on K659. (Angel Montesdeoca, private message, 20140124) The locus of Q is a central cubic with center O (dashed blue curve on the figure). K659 is the anticomplement of pK(X141, X2) and the isogonal transform of pK(X32, X32). 
