Further notes, mainly on cubics of type cK0

Notation

T(P) is the tripolar of P.
C(P) is the circumconic with perspector P.
I(P) is the inconic with perspector P.

For fixed F, P* is the F2-isoconjugate of P.

Theorem 1
If K = cK0(#F,R), then the pivotal conic is the dual of C(R) with respect to C(F).

Outline proof - calculations in apr30.mws
We have an explicit equation for C*(F,R), the dual of the pivotal conic.
We can write down the equation of the dual of U = [u,v,w] with respect to C(F).
If U is on C(R) - so u = -r/(s/v+t/w) - then we can find the condition that the
coefficients - the coordinates of the dual - satisfy the equation for C*(F,R).
This is essentially (r/f+s/g+t/h)(f/u+g/v+h/w) = 0.
As K is a cK0, the first factor vanishes, so the point is on C*(F,R), and hence
the polar is tangent to the pivotal conic.

Note that the proof shows that for the intersections of C(F) and C(R), we also get
tangents to the pivotal conic. The vertices give the sides of the anticevian of F.
We get one new tangent as the tangent to C(F) at the fourth intersection with C(R).

We now have four tangents to the pivotal conic. For K = cK0(#F,R), we can identify
the contact point geometrically.

Notation
Let F1 be the fourth intersection of C(F) and C(R). The first barycentric is f/(s/g-t/h).

Theorerm 2
(1) If K = cK0(#F,R), then the tangent to C(F) at F1 meets C(R) again at a point
of the pivotal conic.
(2) The perspector of the pivotal conic as an inconic of the anticevian triangle of
is the R-Ceva conjugate of F. It lies on the line FF1.

Proof
Theorem 1 shows that the tangent is also tangent to the pivotal conic.
The contact point is the dual of the tangent to C*(F,R). A Maple calculation shows
that the point derived from F1 is on C(R) if and only if K is a cK0. Then the point
has first barycentric r/(s/g-t/h) which clearly lies on C(R) and the tangent to C(F)
at F1.
The perspector can be calculated from the tangents to C(F) at the vertices. Maple
gives the required formula for the coordinates. It is routine to check that it is on
the line FF1.

There is a better way - via the inconic I(1/R). See may17

Theorem 3
K = cK(#F,R) is of type cK0 if and only if the tripolar of F is tangent to the pivotal
conic. In this case the contact point is R.

Proof
This is simply the condition that the dual of the tripolar - the isotomic conjugate of F -
lies on the dual of the pivotal conic.
The dual of rfx2 + sgy2 + thz2 - (tg+sh)yz - (rh+tf)zx - (sf+rg)xy = 0.
This contains 1/F, with tangent rx + sy + tz = 0.

We now have five tangents to the pivotal conic - the tangents to C(F) at all four
intersections with C(R) and the tripolar of F.

Theorem 4
If K = cK0(#F,R), then R*, the tripole of L(F,R), lies on C(F) and on FF1.

Proof
The tripole has first barycentric f2/r, so is on C(F).
The second assertion is a routine calculation.

We say that the tripole is the F-antipode of F1 on C(F).
Note that the nodal tangents are also tripolars of F-antipodes as F is on L(F,R)

Theorem 5
For K = cK0(#F,R), the nodal tangents are the tangents from F to the inconic I(R*).
The contact points are the intersections of the inconic with the tangent to C(F) at R*.
The inconic I(R*) touches T(F) at R.

Proof
We know that the nodal tangents are the tripolars of the intersections of C(F) and T(R*).
Taking isotomic conjugates, the tangents are the duals of the intersections of T(1/F)
and C(1/R*), where 1/P denotes the isotomic conjugate of P.
Taking duals, the nodal tangents are also the tangents from F - the dual of T(1/F) - and
I(R*) - the dual of C(1/R*).
The contact points are the intersections of the inconic with the polar of F for the inconic.
A simple calculation shows that this is the stated line. It has equation
                   (r2/f3)x+(s2/g3)y + (t2/h3)z = 0.
The last part is easily seen by noting that 1/F is on the dual C(1/R*) with tangent
                   rx + sy + tz = 0.

We also have Theorem 3, so I(R*) touches the pivotal conic at R.

We have a strange concurrence which may be useful

Theorem 6
The tangents T3, T4 to C(F) at F1 and R* are concurrent with T(F) and T(R).

Proof
The line F1R* passes through F by Theorem 4. Hence the tangents at F1 and R* meet on
the polar of F. This is T(F). It is easy to check that the point has first barycentric given by
r(s/g-t/h). The point clearly lies on T(R).

The following sums up and extends some earlier observations.

Theorem 7
For K = cK0(#F,R), let Δ be the triangle formed by T(F) and the nodal tangents of K.
Then Δ has inconics the pivotal conic and I(R*) and circumconic C(R).
For each inconic, the perspector with respect to Δ lies on FR.
For the circumconic, the perspector lies on FR*.

Proof
The fact that they are inconics is obvious from earlier results. The inconics touch T(F) at R.
It follows that the perspectors lie on FR as F is the vertex opposite T(F).
The nodal tangents meet C(R) on T(F), so Δ has C(R) as circumconic.
It is easy to verify that the tangents at the vertices other than F meet at U, the pole of T(F).
This is the R2-isoconjugate of F. It is easy to check that U lies on FR*. Hence the perspector
lies on FR*. U has first barycentric r2/f.

Note. U is also the F-Ceva conjugate of R - the point which arose in connection with the view
of K as the F-Hirst inverse of C(R).

Strangely, the polar of the point U in Theorem 7 with respect to C(F) passes through the
point in Theorem 6. This follows since U is on FR*, the polar of the point in Theorem 6.
This polar is also the polar of F with respect to the pivotal conic (see may7.mws).
The equation is ((t/h)2+(s/g)2)x/f + .... = 0.

We have quite a number of coincidences, which we now summarize.For brevity, we describe
a point by its first barycentric coordinate, and a line by the coefficient of x in the equation.

The tangent to C(F) at F1 is
T3 : (s/g-t/h)2/f,
The tangent to C(F) at R* is
T4 : r2/f3,
The coefficients of T3 satisfy the equation of the dual of the pivotal conic.
Thus T3 is also a tangent to the pivotal conic as well as C(F).
The tangent to this dual at this point is
T5 : r/(s/g-t/h), (see may10.mws),
It follows that the coefficients of T5 give the point on the pivotal conic.
F4 = r/(s/g-t/h)
This clearly lies on C(R), and on FR.

We have the fact that the tangent to C(F) at F1 is also tangent to the pivotal conic.
The point of contact with the latter lies on C(R).

T3, T4 meet at the point
F3 = r(s/g-t/h) (see Theorem 6).
This lies on T(F) and T(R).
The polar of F with respect to the pivotal conic is
P : ((s/g)2+(t/h)2)/f (see may7.mws).
It is easy to check that P meets T(F) at F3.
It is also easy to check that P is also the polar of U= r2/f with respect to C(F).

Note that F3 lies on five significant lines.

Notes on general cK(#F,R)

Theorem 8
The nodal tangents of K = cK(#F,R) are the tripolars of the intersections of C(F) and the
polar of R with respect to C(F) - the line x(s/g+t/h)/f + y(t/h+r/f)/g + z(r/f+s/g)/h = 0.

Proof
This follows from the fact that the nodal tangents are the tripolars of the contact points
of the tangents from R to C(F). The line is the polar of R.

The polar of F with respect to the dual of the pivotal conic is x(2f/r+g/s+h/t)/f + ..... = 0.

Some special cubics of type cK0

Suppse that TG(U) denotes the tripolar centroid of U. If U = [u,v,w], then TG(U)
has first barycentric u(v-w)(v+w-2u).

The circumconic C(TG(U)) has infinite points given by u(v-w) and (v+w-2u).
It also contains U and at least five other points with simple coordinates
including R1,R2 with first barycentrics u(v+w-2u), u/(v-w)2.

Note that R1 lies on T(F) as well as T1

Theorem 9
For F = [f,g,h], let K = cK0(#F,TG(F)). Then the nodal tangents T1,T2 are the tripolars
of the points P1, P2 given by f/(g-h), f/(g+h-2f).

Proof
It is trivial to check that these lie on C(F) and L(F,TG(F)) since the latter has tripole
with first barycentric f/(g-h)(g+h-2f).

Theorem 10
For F = [f,g,h], let K = cK0(#F,TG(F)), {R1,P1,F1} and {R1,P2,R*} are collinear triples.
Proof
These are just simple algebraic checks.

The Parry Cubic

K(K) = cK0(#K,TG(K)).

The tripolar of K is the Lemoine Axis. This is then a tangent to the pivotal conic.

The conic C(TG(K)) contains K, R1 = X(187), R2 = X(249), X(512), X(524) by standard theory.
By looking at isotomic and isogonal conjugates, it also contains
X(598) - the perspector of the Lemoine Inellipse,
X(843) = F1 - the fourth intersection with C(K) - the circumcircle.

Thus K(K) contains K and X(843).

L(K,TG(K)) has tripole R* = X(691), the K-antipode of X(843) on the circumcircle.

The nodal tangents have tripoles P1 = X(110), P2 = X(111), which are the intersections of the
Circumcircle and the Parry Circle.

The nodal tangents are then
T1, the Brocard Axis,
T2, the line through K parallel to the Lemoine Axis.

The pivotal conic PC is the inconic of the tangential triangle touching the Brocard and Lemoine
Axes, as well as T2.

R1,P1,F1 = X(187),X(110),X(843) are collinear, and X(352) is also on this line.
R1,P2,R* = X(187),X(111),X(691) are collinear, and X(23) is also on this line.
Of course X(352) and X(23) are on the Parry Circle.

PC is the dual of C(R) with respect to C(F),

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