too complicated to be written here. Click on the link to download a text file. X(4), X(74), X(265), X(14524) points at infinity of the McCay cubic K003 points of K668 on the circumcircle among them Q1, Q2 on the perpendicular bisector of OH S = X(14524) = X(5)X(523) /\ X(4)X(265) two common points of the line X(3)X(265) and the rectangular circum-hyperbola passing through X(186). These points are isogonal conjugates and lie on the McCay cubic K003 and on K668
 K669 is a stelloidal non-pivotal cubic with radial center the point X = X(14644) = homothetic of X(265) under h(X5, 1/3). K669 is spK(X3, X125) as in CL055. See the analogous K930. *** The union of the line at infinity with the Jerabek hyperbola and K669 generate a pencil of stelloids whose basepoints are A, B, C, X(4), X(74), X(265) and the points at infinity of the McCay cubic K003. Let K(Y) be the member of the pencil passing through the finite point Y on the line X(74)X(265). K(Y) is the stelloid spK(X(3), Q) where Q is the midpoint of X(3), Y. The radial center X is the homothetic of Y under h(X5, 1/3). K(Y) meets the Euler line at X(4) and two other points E1, E2 which are inverse in the circumcircle (O) of ABC. K(Y) meets (O) at A, B, C, X(74) and two other points O1, O2 which are symmetric about the Euler line. Note that the line O1O2 meets K(Y) again at Z on the line X(4)X(265). These points also lie on pK(X6, Y). K(Y) meets K003 at six finite points namely A, B, C, X(4) and two other points M1, M2 on the line X(3)Y. These points are isogonal conjugates and also lie on pK(X6, Y). Remark : in general, if E1 and E2 are real then O1, O2 are imaginary conjugates and vice versa. The four points are all real if and only if K(Y) contains X(1113) or X(1114) but, in each case, they all coincide and the cubic has a node at this point. See figure below.
 K(Y) with real points E1, E2 K(Y) with real points O1, O2