nK of Hirst type
We saw earlier that a cK of Hirst type was actually an nK0, and so a cK0.
In fact an nK of Hirst type must be a cK0.
Suppose that K is a cubic of nK type, and is the F-Hirst inverse of a circumconic R.
Let F', F" be the points [f,θg,θ2h], with θ;3 =1, θ ≠ 1. F is on the tripolars of F', F".
If F is not on R, each of these tripolars cut R in two points, so we have two double
points on K unless F' or F" lies on R. If F', F" are on R, R = F and the F-Hirst inverse
decomposes as F and the tripolar of F. Thus F or one of F', F" is on R.
For any point P, we define analagously the points P', P". We shall write Q for the
circumconic with perspector Q.
A tedious Maple calculation proves the following
Theorem
If K =nK(W,S) is the F-Hirst inverse of R with perspector R, then we have
K = cK0(F,R), cK0(F',R") or cK0(F",R').
Note that we also have
cK0(F',R) is the F-Hirst inverse of R', cK0(F",R) the F-Hirst inverse of R".
Outline of calculation
Recall that F-Hirst inversion is a bijection on the complement of the tripolars of F, F', F".
We call the points of this complement good points. Suppose we take a good point P on
the conic R. Let Q be the F-Hirst inverse. Then the coordinates of F(Q) - the F-Hirst
inverse - satisfy the equation of K. This is quartic. Since K has infinitely many points,
the quartic must factorize into the equation of K and of a line L. Unless L is the tripolar
of one of F, F', F", it contains infinity many good points. These must map back to R.
But the image of a line is a conic (not a circumconic), so we have just four intersections
with R. This is a contradiction. Thus L is one of the tripolars.
This allows us to write down both factors. The Kxyz term gives rise to terms with all of
x,y,z. Looking at the terms with a cubic identifies the root of the resulting cubic.
Eventually, we arrive at equations which are consistent if and only if K = 0.
We also identify the node as the point F, F', F" related to L. When the node is F, the root
is equal to S.
Of course much of this could be obtained by geometry. Gibert's list contains two explicit
examples of Hirst type. K040, K185. It also contains two others K137, K147. Take any of
these as K = cK(F,R), with F the node, R the root. Then R is on the tripolar of F. Now take
another point E. There is a projective transformation ρ fixing A,B,C and sending F to E.
The Hirst inverse is a projective map, and preserves isoconjugation and perspectors and
tripolars. It sends R to Q on the tripolar of E. It sends the F-Hirst inverses to the E-Hirst
inverse and R to Q. This therefore sends cK(F,R) to cK(G,Q) as the G-inverse of Q. We
could use the Theorem to identify the image as cK(G,Q), but could prove most of the
theorem by observing that the node and root properties are projective. This approach
does not reveal the other possibilities, or that an F-Hirst conjugate must be a cK0.
Suppose that we start with ΔABC, and circumconics with perspectors R and F, with F on R.
Then we have a Hirst type cubic cK0(F,R). The projective map preserving R, interchanging
A,B and mapping D, the fourth intersection of R and F, is a (Fregier) involution. This must
preserve all conics through A, B, C, D, such as F. Take ΔABD. Then the perspector of F
with respect to ΔABD lies on R. Then we have a cubic of Hirst type for ΔABD. Similarly for
ΔACD and ΔBCD. Of course here the "poles" of the Hirst inversion is different in each case,
being the perspectors relative to the different triangles. The two conics are R, F in every
case. This shows the dependence of the "pole" of inversion.
Conjecture
Given four points, with no three collinear, and a conic R through these, then there is one
conic of the family of conics through all fourwhose perspector lies on the tripolar of the
perspector of R with respect to three of the points.
Then, given a conic R through four points, we have just four cubics of Hirst type of the
form cK0(F,R).