letter 1
I have been looking at C031
I came to X and X' in a different way.
We know that if TG(U) = P, then there is a second point V with TG(V) = P.
Indeed, if U = (u,v,w), V = ((2u-v-w)/(2vw-uv-uw),..,..).
Let us define the map m taking U to V.
There must be only one point on GR with tripolar centroid on GR
since such a point is on the cubic.
It turns out be your X'.
Then X' is m(T) and is the intersection of GR with TT'.
By the way, the conic C(TG(T)) passes through T and X' of course,
but also through R.
It touches the cubic at T.
I got X as the isotome of X', but also as m(T').
In barycentrics X' is not very nice. If R = (r,s,t), X' = (r^2+2st)/(s-t).
In general, for any U on the cubic, C(TG(U)) passes through T as the
perspector lies on GR.
It also contains U and the vertices.
The sixth intersection with the cubic must be m(U).
We know that for any U, m(U') = m(U)' (see the above formula), so that
we get a set of four points U,U',m(U),m(U') closed under m and isotome.
The tripolar centroids are reflections in G.
For U on the cubic
Um(U) passes through T' - this characterises the cubic.
We have a quadrilateral with nice intersections ; see mates.mws
Um(U) and U'm(U') meet at T' i.e. are parallel,
Um(U') and U'm(U) meet at T,
UU' and m(U)m(U') meet on GR.
Perhaps this might be useful.
At the moment, there are no examples in ETC with U = m(V).
I have written about the case cK0(#G,R)
but the theory works equally well for cK0(#F,R) in general
with T(F) instead of the line at infinity.
The definitions of TG and m need to be altered of course,
but the theory is entirely projective.
letter 2
I find that in the work tX appears more often than X, so I write tX = R = [r,s,t].
Then cK(#G,R) is the locus of points P such that
(1) C(TtR(P)) touches T(R),
or such that
(2) TtR(P) lies on I(R) - the inconic perspector R.
(2) looks a bit like C029, but (1) is more useful.
If U = [u,v,w] is on T(R) then the circumconic touching
T(R) at U has perspector [u^2/r,.,.].
This conic meets GU
at f(U) = [u/r(v-w),.,.], which then lies on cK(#G,R).
There are two obvious points on T(R) the infinite point [s-t,.,.]
and the intersection with the dual of R [r(s^2-t^2),.,.].
These lead to points on the cubic [(s-t)/(r^2-st),.,.]
and [(s-t)/(2st-rs-rt),.,.], and their isotomes.
For R at infinity, these are (in C029 notation) T,T',X,X'
Problem - What are these points in general?
If U+ = [r(v-w)], then U+ is on T(R) and we find that
f(U+) = tf(U).
When U = U+, the point is G - we have a nodal tangent GU.
The condition U = U+ gives U on T(R) and the diagonal conic
D(R) x^2/r+y^2/s+z^2/t = 0.
Indeed, the nodal tangents
are the tangents from G to D(R) as the polar is T(R).
Finally, after some examples (see below) I found that
The points U,V on T(R) such that GU, GV are nodal tangents
to cK(#G,R) are the unique pair of G-Ceva conjugates on T(R).
Problem - what do the conics C(U),C(V) have to do with cK(#G,R)?
Examples
cK(#G,X(110)) - T(R) = OK, the Brocard axis
Nodal tangents are the GO,the Euler Line, and GK.
The cubic contains X(97),X(324), X(251) and its isotome.
cK(#G,X(100)) - T(R) = IM, M = X(9)
Nodal tangents are GI, GM.
The cubic contains X(81) and X(321).
Generalisation (by projective transformation)
The points U,V on T(R) such that FU, FV are nodal tangents to cK(#F,R) are the unique pair of F-Ceva conjugates on T(R).
The diagonal conic is D(Barycentric product FR).
Example (isogonal)
cK(#I,X(101)) - T(R) = KX(55)
Nodal tangents are IK, IX(55).
The cubic contains X(10), X(58).