Working notes on cK0

see also text

We use the generalised tripolar centroid to discuss the family cK0(#F,R).
Our theory here is entirely projective, so we consider just the case F = G
in our calculations when this is more convenient.

To translate

Replace point [F(u,v,w)] by [fF(u/f,v/h,w/h)]
Replace line ΣC(u,v,w)x = 0 by C(u/f,v/g,w/h)(x/f) = 0.

Definition - special case of TFR
For U[u,v,w] ≠ F[f,g,h], TFF(U) = [u(v/g-w/h)(2u/f-v/g-w/h)].

Result 1
TFF(U) is the perspector of the circumconic through U and FUnT(F).

Of course, TGG(U) is the tripolar centroid TG(U).

In general there are two points with same value of TFF.
The circumconic C(TFF(U)) will cut T(F) twice.
Once on FU, the other at W, say. Then FW cuts C(TFF(U)) again at a point V.
Clearly the circumconic is that through V and FVnT(F), so has perspector TFF(V).
We can specify V in terms of U.

Definition
For U ≠ F, MFF(U) = [f(2u/f-v/g-w/h)/(2fvw-gwu-huv)].

Result 2 see june2.mws
TFF(MFF(U)) = TFF(U).
MFF has order 2.

This is related to cubics :

Result 3 also june2.mws
cK(#F,F) = {P : MFF(P) = P} - then C(TFF(P)) touches T(F) - TFF(P) is on I(F)
For R on T(F), cK0(#F,R) = {P : TFF(P) is on FR}.

A simple observation on isoconjugates fU = [f2/u] is our

Result 4
MFF(fU) = fMFF(U).

Consider cK0(#F,R). Given a point P on FR, there are two points on the cubic
such that TFF(X) = P. There seems to be no sensible way to use this to parametrize
the cubic using FR.

For the moment, we work with F = G, and consider cK0(#G,R), so R = [r,s,t], with r+s+t = 0.

Suppose U = [u,v,w] is on T(G), so u+v+w = 0. Now GU meets the cubic at G and one other point V.
Then we note that U is on C(P), where P = [u(ur+vt+ws)] = [u(v(t-r)-w(r-s)] = [u(s(w-v)-t(u-v)].
Using the second version, we see that P is on GR, so is TFF(W) for some W on the cubic.
In fact
W = [[(s(w-u)-t(u-v))/(v-w)]
since this is on GU and C(P).
And W is on the cubic.

Thus we have a geometrical parametrization in terms of T(G) :

U on T(G) maps to the finite intersection of GU and the circumconic with perspector T(U)nGR.

Note that the perspector is that on GR whose circumconic contains U.

Now MFF(W) comes from the intersection of C(P) and T(G) other than U.
This is U' = [ur+vt+ws] since it is clearly on T(G) and C(P) (as u+v+w = 0).

Now an unpleasant calculation gives MFF(W) as [u/(vs-wt)].
It helps to observe that, from its definition, (U')' = U as there is only one circumconic
through U and U'

By inspection, we see that tW - also on the cubic - comes from X' with X = [v-w].

(*) This is V = [r(v-w)+t(w-u)+s(u-v)] = [vs-tw].

Geometrically, this is the intersection of T(G) and the line through tU and tR.

We will get a nodal tangent (i.e. W = tW) when U = [vs-wt] .
The condition for this reduces to U on D(tR) : rx2+sy2+tz2 = 0.
This is a circumconic of the anticevian of G, passing through G.
The tangent at G is T(tR), which identifies ths conic.
By direct calculation, the perspector relative to the anticevian triangle is R.
Now T(G) is the polar of tR for D(tR), and tUtR the polar of U, so V (see(*))
is the pole of UtR.

Either by brute force or by observing that if U is on D(tR), so is V =[v-w],
or indeed that the quartet U,U',V,V' has four elements except for the points
on the nodal tangents, we have the following :

Result 5
(1) The infinite points on the nodal tangents of cK0(#G,R) are of the form U =[u] and V = [v-w].
(2) R is the barycentric product U&V, i.e. [r] = [u(v-w)].
(3) cK0(#G,R) contains [s-t]

Comments.
(2) comes from the fact that if [u] is a root of the equations, so is [r/u].
(3) comes by using R as a point on T(G) and forming "(ys-zt)/x" for X = R.
In our language, it comes from [v-w] on T(G). In C031 language this point is T'.
Its isotome - the point T' - comes from R' = [rr+st+ts]
The point X = ((r2+2st)/(s-t)] comes from directly from R.
Its isotome - the point X' - comes from [v-w]' = [vs-wt].

Now the general versions :

Result 6
Suppose that R is on T(F), so we have the cubic K = cK0(#F,R).
(1) There is a map KFR from T(F) to K taking U on T(F) to the intersection (other than U)
of FU and the circumconic with perspector T(U)nFR.
(2) fKFR(U) = KFR(V), where V = [f(vs/g2-wt/h2)] is the intersection of T(F) and the line fUfR.
(3) The nodal tangents of K meet T(F) at its intersections with the diagonal conic given by
rx2/f3+sy2/g3+tz2/h3 = 0. This is the circumconic of the anticevian of F, touching T(fR) at F.
The persector of the conic relative to the anticevian is R.
(4) The point V in (2) is the pole of UfR with respect to the conic in (3).

Result 7
Suppose that R is on T(F), so we have the cubic K = cK0(#F,R).
Suppose also that U is the intersection of T(F) with a nodal tangent.
Let V be the intersection of T(F) with the second nodal tangent.
(1) R = [u(v/g-w/h)] = T(F)nT(U).
(2) V = f(v/g-w/h)] = T(F)nT(fU).
(3) The second nodal tangent is T(fU).
(4) The nodal tangents meet T(R) in the F-Ceva conjugate points
[u2(v/g-w/h)/f] and [u(v/g-w/h)2].
These are TFF(U) and TFF(V).

All are easy to verify. Part (4) is just a computation of the intersections of the nodal tangents
and T(R). They are F-Ceva conjugate by general theory.

Corollary 7.1
Given a line L through F, there is a unique cK0(#F,R) with L as nodal tangent.

One obvious choice is L through F parallel to T(F) - then we have from Result 7(1)

Corollary 7.2
The unique cK0(#F,R) with a nodal tangent parallel to T(F) is ST(F) - the Stothers cubic!

Examples

cK0(#G,X(900))
This arises from Result 7 by taking U = X(514) = [b-c].
Then R = X(900), V = X(519).
The nodal tangents are GX(514) and GI.
The cubic contains X(545) as its T' = [s-t] point - C031 notation.
The G-Ceva conjugates on T(R) are TG(X(514)) = X(1647) and TG(X(519)) - unnamed.

cK0(#G,X(690))
This arises from Result 7 by taking U = X(523).
Then R = X(690), V = X(524).
The nodal tangents are GX(523) and GK.
The cubic contains X(543) as its T' = [s-t] point - C031 notation.
The G-Ceva conjugates on T(R) are TG(X(523)) = X(1648) and TG(X(519)) = X(1649).

Now a fact inspired by Cabri

Result 8
In the notation of Result 5, the intersections of T(F) with the nodal tangents are the points
on T(F) whose tripoles pass through R (i.e. the intersections of T(F) and C(R)).

This is simly proved by verifying that for U,R on T(F) the conditions Σr/u = 0, Σru2/f3 = 0
are equivalent.

We have other proofs of this!

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