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X(6), X(44), X(187), X(524), X(1990), X(3284)
Sa, Sb, Sc extraversions of X(44)
midpoints of ABC
other intersections of the Steiner inellipse and the Thomson cubic K002
more details below
For any root P ≠ G on the Thomson cubic K002, there is a nK0 having a pencil of circular polar conics. Its pole Ω must the barycentric product of P and the infinite point of the line GP. In other words, for any nK0(Ω, P) of this kind, there is a line of points whose polar conics are circles which obviously form a pencil. This line is called the circular line of the cubic. When Ω is X(187), X(1990) we obtain K604, K393 respectively.
The transformation P -> Ω maps the Thomson cubic to Q087.
If P and P* are two isogonal conjugates on K002 (hence collinear with G), the corresponding poles Ω and Ω' have the following properties :
Q087 passes through K which is a triple point with tangents passing through the points Ta, Tb, Tc of K002 on the circumcircle. Note that the tangents at A, B, C are the symmedians.
This tricuspidal quartic is tritangent at the midpoints A', B', C' to the sidelines of ABC.
The cusps are the images of A, B, C under the homothety with center G, ratio 3/2.
The cuspidal tangents are the medians of ABC.
Let T be a point on the Steiner inellipse and (T) the tangent at this point.
S is the isotomic conjugate of the infinite point of (T) and aT is the anticomplement of T.
The parallel at G to the line S-aT meets the Thomson cubic at P, P*.
The parallel at T to this same line contains Ω and Ω' which also lie on (C), the isogonal transform of the line TG.
This latter parallel envelopes the quartic, the point of tangency Q being the reflection about T of its second intersection T' with the Steiner inellipse.
The same construction applied to the reflection T1 of T about G gives another point Q1 on the quartic and the line QQ1 is also a tangent to the quartic.