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Let Ω = p : q : r be a point and K(Ω) the pseudopivotal cubic psK(Ω, X2, X3). See PseudoPivotal Cubics and Poristic Triangles, specially Tables 3 and 4. See also the ThomsonLemoine pencil in Table 50. The equation of K(Ω) is : ∑ p b^2 c^2 y z ((b^2  c^2) x + a^2 (y  z)) = 0, showing that all these cubics form a net which can be generated by three independent members. It is convenient to choose K(A), K(B), K(C) since these are very simple. Indeed, K(A) is the decomposed cubic, the union of the sidelines AB, AC and the perpendicular bisector of BC. Each cubic K(Ω) contains seven fixed points namely A, B, C, X3 and the midpoints of ABC. Recall that the tangents at A, B, C concur at the pseudoisopivot Ω. K(Ω) is a pK when it contains G in which case Ω must lie on the Brocard axis. See below. K(Ω) is invariant under two simple transformations, one being the inverse of the other : T1 : M = x : y : z → p y z (x + y  z) (x  y + z) : : , T2 : M = x : y : z → p y z ( p y z + q x z + r x y) : : . *** Other points on K(Ω) : • K(Ω) meets the line at infinity at the same points as pK(X6, atgΩ). The six remaining common points lie on the circumconic with perspector the X32isoconjugate of the tripole of the line ΩX6. This latter conic passes through gtgΩ. • K(Ω) meets the circumcircle at the same points as pK(X6, tgΩ). The three remaining common points lie on the line which is the isogonal transform of the conic above and one of them is the pivot tgΩ. • Apart tgΩ, K(Ω) also contains the Ωcrossconjugate of X(6) which is gatgΩ. Note that the three cubics coincide into K002 when Ω = X6. Now, if mΩ denotes the midpoint of atgΩ, tgΩ (which is actually ctgΩ and also the isogonal conjugate of the cevapoint of X6 and Ω), it can be easily verified that K(Ω) is spK(atgΩ, mΩ) as in CL055 with all the consequences mentioned there. *** Isogonal transform K(Ω)* of K(Ω) K(Ω)* is psK(X6 x gΩ, gΩ, X4) and also spK(tgΩ, ctgΩ). Its pseudoisopivot is X(6) hence the tangents at A, B, C are the symmedians. When they are distinct i.e. Ω ≠ X6, these cubics K(Ω) and K(Ω)* generate a pencil of circumcubics which are all spK(P, ctgΩ) for some point P on the line L(Ω) passing through X(2), tgΩ, ctgΩ, atgΩ. Every cubic of this pencil passes through A, B, C, the four foci of the inconic with center ctgΩ and perspector gΩ, and two (not always real) isogonal conjugate points on the line G, ctgΩ hence also on the circumconic passing through K and gctgΩ (in fact, the cevapoint of K and Ω). The pencil contains : • one and only one pK, namely pK(X6, ctgΩ), • a focal isogonal cubic with singular focus F on (O). F is the isogonal conjugate of the infinite point of the line L(Ω), obviously also on the cubic which is a nK with root the infinite point of the trilinear polar of gΩ. • an equilateral cubic if and only if Ω lies on the Lemoine axis (but these cubics decompose into the line at infinity and a circumconic) or on the trilinear polar L(X112) of X(112) in which case the cubic is a McCay stelloid. K(Ω) is a member of the pencil studied in Table 50. L(X112) passes through X(i) for i = 6, 25, 51, 154, 159, 161, 184, 206, 232, 1194, 1474, 1495, 1660, 1843, 1915, 1971, 1974, etc. *** More generally, the isoconjugation with pole P transforms K(Ω) into psK(P^2÷Ω, P÷Ω, P÷O) where ÷ denotes a barycentric quotient. In particular, the isotomic transforms of K(Ω) is psK(tΩ, tΩ, X264). Anticomplement aK(Ω) of K(Ω) aK(Ω) is obviously a circumcubic since K(Ω) contains the midpoints of ABC. It is a pK when Ω lies on the Brocard axis (recall that, in this case, K(Ω) is also a pK, see below) and a psK when Ω lies on the bicevian conic C(X3, X6). In this case, the pseudopole lies on K555, the pseudopivot lies on the Steiner ellipse, the pseudoisopivot lies on the line at infinity. Hence, the tangents at A, B, C are parallel. Recall that C(X3, X6) is the locus of the X(3)Ceva conjugate of every point on the Lemoine axis, and equivalently, the locus of the X(6)Ceva conjugate of every point on the trilinear polar of X(3), a line passing through X(i) for i = 520, 647, 652, 684, 686, 852, 1459, 1636, 3049, etc. Note that C(X3, X6) is the barycentric product of O and the nine points circle. It contains X(i) for i = 3269, 7117, 9475, 15166, 15167, 20728, 20830, 20975, 22084, 22428, 47405 → 47424. The points X(15166) and X(15167) lie on the Brocard axis and on the Steiner inellipse which is bitangent to C(X3, X6) at these points. The common tangents meet at X(647). 



Special cubics K(Ω) Pivotal cubics K(Ω) is a pK if and only if it contains X2 hence Ω must lie on the Brocard axis. In this case, these cubics form a pencil and each cubic passes through the seven fixed points above and also X2, X6. Obviously, K(Ω) contains the (real or not) square roots of Ω on the Stammler rectangular hyperbola. gΩ lies on the Kiepert hyperbola and K(Ω) also contains the foci of the inconic with perspector gΩ and center ctgΩ. The following table gives a list of all CTC cubics and also a selection of other cubics that pass through at least seven ETC centers (2, 3, 6 are not repeated) and/or some other interesting points. 





Circular cubics K(Ω) is circular if and only if Ω lies on the Lemoine axis. In this case, these cubics also form a pencil and each cubic meets the line at infinity again at a real point M = tgΩ and the circumcircle again at gM = gtgΩ. The real asymptote is the Simson line of the antipode of gM on (O) hence it envelopes the Steiner deltoid H3 and it meets K(Ω) again on Q011. The singular focus F of K(Ω) lies on C(X3, R/2) since it is the midpoint of X3gM. The only pK of the pencil is K043 = K(X187) with focus X(14650). There are three focal cubics in this pencil and one of them is always real. They are obtained when Ω is the barycentric product of X6 and one of the infinite points of the Napoleon cubic K005 or, equivalently, the common points of the Lemoine axis and pK(X1501, X51). The singular foci are the midpoints of the segments joining X3 and the common points of (O) and K005. These three focal cubics are actually central cubics, symmetric about their singular focus. 

Circular K(Ω) 
Focal K(Ω) 

The following table (computed by Peter Moses) gives a list of all CTC circular cubics and also a selection of other cubics together with the points M, F mentioned above. 





Other remarkable cubics K(Ω) is equilateral if and only if Ω = X(51) corresponding to K026. K(Ω) is a psK+ if and only if Ω lies on a complicated cubic passing through X(51), X(154) corresponding to the cubics K026, K426 which are actually both central cubics. K(Ω) is an unicursal cubic if and only if Ω lies on a very complicated sextic passing through X(55), X(184), X(8041) corresponding to the cubics K259, K009, K1068. The table below presents other listed cubics and a selection of K(Ω) with at least seven ETC centers. Some additional points are given : (∞)Knnn and (O)Knnn denote the infinite points and the points on the circumcircle of Knnn. 



Remark : for any Ω on the line passing through X(6) and X(41), K(Ω) passes through X(9) and X(57) just like K002.

