# Proof of Thomson cubic pivotal property without coordinates

The Thomson cubic is defined as the cubic going through A,B,C, the three side midpoints, the three excenters. Is there a way to prove its pivotal property (any two isogonal conjugates on it have a line through the centroid) with purely synthetic geometry techniques and not using barycentric coordinates?

A more meta-question: how to prove things about triangle cubics with pure Euclidean geometry?

• Wikipedia's "Cubic plane curve" entry defines the Thomson cubic by the pivotal property: "the locus of a point [whose isogonal conjugate] lies on the line [through the pt and the centroid]." So, maybe there's nothing to show. :) Then again, the "Thomson cubic" entry gives a very different defn; and the Catalogue of Cubics "K002" entry lists two dozen equivalent locus defns. What's the appropriate starting place? (cont'd)
– Blue
Commented Jun 28 at 21:01
• (continuing) Also, showing equivalences among the locus defns typically involves showing that the (barycentric/trilinear/etc) coords of their points satisfy a given cubic relation, so using those defns may not officially count as "pure Euclidean geometry". Presumably, one could show that a locus is generically "cubic" in nature, and that it shares enough pts in common with K002 to be the same curve. But this requires a suitable "synthetic" defn of a generic cubic, which can be trickier than just using coords. See, eg, this Math.SE question.
– Blue
Commented Jun 28 at 21:01
• By the way, the Euclid mailing list at Groups.io is frequented by regular contributors to the Encyclopedia of Triangle Centers and the Catalog of Triangle Cubics. They will likely have useful insights for you. ... Good luck!
– Blue
Commented Jun 29 at 10:56