Notes May 17 2004

Random notes on work in progress

We work with a cubic K = cK0(#F,R) - so R on T(F).

1. Hirst Summary - see Maple may17.mws, may17b.mws

Let hF(X) be the F-Hirst inverse of X. Note that it is of order two.

Definition
U = r2/f

The Maple shows that
the line hF(X)hF(X*) passes through U = r2/f
if and only if X is on a union of two cubics, one of which is cK0(#F,R).

We also see that hF(X) is on K if and only if X is on C(R) by checking
that hF(X) is on K only for X on C(R) or three lines which map to F.
Equally we can check that hF(X) is on C(R) only for X on K or on the
line T(F) which maps to F. F is on C(R).

Thinking of the inverse map,
isoconjugates on K correspond to points on C(R) which are U-antipodes.

Maple may17b.mws shows that for a cK0, the condition that a line which
is through F meets T(F) at a point of C(R) is the same as that for the line
to be a nodal tangent.

2. Dual results see Maple may17c.mws.

The first few steps of Maple show that in general the coefficients of the
polar of X on C(R) lies on the dual of the pivotal conic if and only if R is
on T(F) or X is also on C(F). Then, taking duals, the polar is tangent to
PC, the pivotal conic

For a cK0 we have PC is the dual of C(R) with respect to C(F).

For any cK, we get four tangents - at F1 and the vertices A,B,C of the
reference triangle. The latter are the sides of the anticevian triangle of F.
The former gives us the polar of F1 = f/(s/g-t/h). Taking the derivatives
in the dual equation, using the coefficients of the polar, we can get the
tangent to the dual. The coefficients are the coordinates of the point of
contact with the PC. In general, we get f(2r/f+s/g+t/h)/(s/g-t/h).

Of course, the tangent is the dual of F1 - so is the tangent to C(F) at F1.
The equation is then

Definition
T3 : Σx(s/g-t/h)2/f

For a cK0, the contact reduces to r/(s/g-t/h) - which lies on C(R) and on FR.
Thus the point is the second intersection of C(R) and FR.

The tangent meets T(F) at r(s/g-t/h) this is the intersection of T(F),T(R).

Definitions
F3 = T(F)nT(R) = r(s/g-t/h).
F4 = C(R)nFR = r/(s/g-t/h)

3. The second tangent from F1 to PC see may17d.mws

This of course is F1F2 as F2 = F1*, and XX* is always tangent to PC.

For a cK0, this is the tangent to C(R) at F1 i.e. T4 : Σr(s/g-t/h)2/f2 = 0

The contact is computed as in 2 as the coefficients of the tangent to
the dual of the PC at the point given by coefficients of T4.

We get contact f((s/g)2+(t/h)2)/(s/g-t/h).

This is hardly worth naming.

4. The tangents from F to the PC see may17e.mws

The tangents are of course the nodal tangents.

The polar of F with respect to the PC is

P : Σx((t/h)2+(s/g)2)/f = 0.

It is also the polar of U with respect to C(F).

4. The significance of the inconic I(1/R) see Bernard Gibert's C031.

Notation
For a vector or matrix P, P' is the transpose of P.
For a square matrix, M- is the adjoint of M.
A conic may be written as X'MX = 0, where M is a symmetric matrix.
The conic ic a circumconic if M has zeros on the main diagonal.

Definition
For a symmetric matrix M, c(M) is the conic X'MX = 0.

Facts
The dual of c(M) with respect to c(N) is c(N'M-N).
The absolute dual (i.e. with respect to c(I)) of c(M) is c(M-).
If c(M) = C(R), the absolute dual is I(1/R).
In our context, say C(F) = c(N), C(R) = c(M).
The pivotal conic PC is then x'NM-Nx = 0.
Thus for X on PC, NX is on c(M-) = I(1/R).

The map fF: X to NX takes X to the isotomic conjugate of the cevapoint of F,X.
Hence the inverse is the map gF taking X to the (1/X)-Ceva conjugate of F.
In coordinates, gF maps (x,y,z) to (f(-fx+gy+hz),g(fx-gy+hz),h(fx+gy-hz)).

Theorem
For cK0(#F,R), the pivotal conic is the image of I(1/R) under gF.

If F = G, gF(X) is the anticomplement of X.

If we apply this to 1/R the perspector of I(1/R), we get

Theorem
For a cK0(#F,R) the perspector of PC as inconic of the anticevian triangle of F
is the R-Ceva conjugate of F.
It is collinear with F,R* and F1.
It lies on the line P in 3 (this needs Maple).

As defined this is f(-fst+grt+hrs) = f((r/f)2+(s/g)(t/h)).

5. Generalized tripolar centroid. see Maple may19.mws

Definition
For F = (f,g,h), TF(X), the tripolar F-centre of X = (x,y,z), X ≠ F is the point
with first barycentric coordinate x(y/g-z/h)(y/g+z/h-2x/f).

When F = G, we get the usual tripolar centroid TF(X).

By inspection, C(TF(X)) meets T(F) at "x(y/g-z/h)" and "f(y/g+z/h-2x/f)".
C(TF(X)) is the circumconic through X and f(y/g+z/h-2x/f) , the intersection
of T(F) and FX.

"x(y/g-z/h)" and "f(y/g+z/h-2x/f)" coincide precisely when X is on cK(#F,F).
Thus:

Theorem
cK(#F,F) is the locus of points for which C(TF(X)) touches T(F).

The classic example is K015 (F = G) the Tucker nodal cubic.

By routine calculation, if R = (r,s,t), then TF(X) is on FR when X lies on the
cubic cK0(#F,S), where S is f(2r/f-s/g-t/h). Obviously we may replace R
by any other point on FR. We might as well choose R on T(F). Then S = R.

Theorem
For F = (f,g,h), R = (r,s,t) on T(F),
cK0(#F,R) is the locus of X such that TF(X) lies on FR.

Of course, cK0(#F,R) must have R on T(F), so all cK0 occur in this way.

As a special case, take FR the line through F parallel to T(F), so R = f(g-h).
Examples include K040 (F = I), K222 (F = K).

A point Y can be written as TF(X) only if C(Y) has real intersections with T(F).
This corresponds to Y not within I(F). Y on I(F) come from points on cK(#F,F).

Note that, in general, a point Y can be described as TF(X) in two ways.
Solving TF(Y) = TF(X), we get Y = X or Y = f(z/h+y/g-2x/f)/(h/z+g/y-2f/x).

In our case, each point on FR arises from two points on the cubic, related
as above.

Given a point Y which is T(Z) for some Z, we can recover the possible Z.
Draw C(Y). This meets T(F) at two points Q1, Q2. Starting from Q1, let R1
be the other intersection of FQ1 and C(Y). Then Y = TF(R1). We get the
second possibility from Q2.

This is checked by assuming we have Y = TF(X).
Then take Q1 = x(y/g-z/h).
We find (by Maple) the points of intersection of FQ1 and C(Y).
They turn out to be Q1 (of course) and the point Z other than X
with TF(Z) = Y.

Note that the point must be on C(Y) since Z is on C(TF(Z)) for all Z.

Construction notes

We construct TF(X) as follows : note that TF(X) is on T(X).
L1 : line joining 1/X, 1/F, coefficients 1/zg-1/yh = x(y/g-z/h)
P1 : tripole of L1,
P2 : 1/P1,
L2 : tripolar of L2 coefficients 1/x(y/g-z/h), contains TF(X)
TF(X) = L2nT(X).

The second part is the construction of KFR = cK0(#F,R).
Take X on FR (outside I(F)) - so X = TF(Y) with Y on KFR.
By the method above,
Draw C(X).
Choose Q as an intersection of T(F) and C(Y).
Let Y be the other intersection of FQ and C(Y).
Then Y is on KFR.

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