cubics by inversion

Several of the cubics in Gibert's list can be described as Hirst inverses of a conic.
The best examples are K040 and K185, though K137 and K147 can be so described.

As defined in ETC, the P-Hirst inverse of a point X is the intersection of the line PX
and the polar of X with respect to the circumconic with perspector P. More generally,
given points P and Q, we can define the (P,Q)-inverse of X as the intersection of the
line XQ and the polar of X with respect to the circumconic with perspector P.

The case where P = Q is special - we have

Theorem 1
The F-Hirst inverse of a circumconic C is a cubic if and only if F is on C.
The cubic is then cK0(#F,R), where R is the perspector of C.

It follows that every cK0 can be described as an F-Hirst inverse of a conic.

In general, the first part is still true, though the relation of the conic and cubic is more
complicated. The cubic, however, is always a cK, though not a cK0.

Theorem 2
Suppose that F=[f,g,h] and P=[p,q,r] are distinct points, F not on the circumconic
with perspector P and that C is the circumconic with perspector K=[k,l,m].
The (P,F)-inverse of the circumconic C is a cubic if and only if F is on C.
The cubic is an nK if and only if K = [f(2p/f-q/g-r/h),g(2q/g-p/f-r/h),h(2r/h-p/f-q/g)].
Then the cubic is cK(#F,U), where U = <fgrp+fhpq+hgp2-3f2qr>

The proof was assisted by Maple. [Personal note: worksheet apr12.mws]

Outline of calculation
Applying the mapping given by (P,F)-inversion to the K-conic C, we get a quartic. This must
split as the cubic and a line. If the line is not the polar of F with respect to the P-conic, then
applying the map again, we get a contradiction as we have infinitely many points on the
K-conic and the conic which is the inverse of the line.

If we now find the condition for points on the polar to satisfy the quadric, we get a quadratic
in x,y,z (the solutions are the intersections of the polar and the K-conic) and the constant
term k/f+l/g+m/h. Since the quadric vanishes infinitely often, this must vanish - F is on C.

Now we can replace k by a combination of l and m. The quadric factorises into the polar of F
and a cubic. The cubic has coefficients which depend on l and m. For the cubic to be an nK,
we require that g2* coefficient of xy2 = h2*coefficient of xz2, and similar conditions on the
coefficients of yz2,yx2, and zx2,zy2. It is remarkable that these are consistent. They lead to
the stated values of l,m and hence k.

It is now routine to check that the xyz term in the cubic is such that it is a cK. The root U can
be read from the equation of the cubic.

Apart from the case F = P, the most obvious case is where F is the centre of the P-conic.
Then we have F =<p(-p+q+r)> and, equally, P = <f(-f+g+h)>. We then get the forms

K = <f(2f-g-h)>, U = <f(-f2+g2+h2-gh)>, or alternatively

K = <p(-p+q+r)(2p2-q2-r2-pr-pq+2qr)>, U = <p(p2+q2+r2-2pq-2pr+qr)>

Notes
(1) In either case, we can verify that <p/(q-r)> lies on the cubic. It is the intersection of
the P- and K-conics.
(2) The point K is the barycentric product of F and the infinite point on GF (unless F = G).
(3) The condition that F is on the K-conic may be restated as "K is on the tripolar of F".
(4) The cubic degenerates when P = F, for then K = P.

Example 1
Take P = X(6), F = X(3) - so the P-conic is the Circumcircle, F its centre.
Then the cubic is cK(#X(3),X(323)), containing X(110).
K is unlisted but is the barycentric product of X(3) and X(30) - it lies on the tripolar of X(3).
The K-conic may be identified as the circumconic through X(3) and X(110).

Example 2
Take P =X(3), F = X(6) - so F is the centre of the circumconic with perspector P.
Then the cubic is cK(#X(6),X(23)), containing X(110) again!
Here K = X(187).

There are two other obvious examples where the P-conic is the circumcircle.

Example 3
Take P = X(6), F = X(1) - so we get an isogonal cubic.
K = X(44), U = <a(a2+ab+ac-3bc)>.
cK(#X(1),U) contains X(100).

Example 4
Take P = X(6), F = X(2) - so we get an isotomic cubic.
K = X(524), U = <a4+a2b2+a2c2-3b2c2>.
cK(#X(2),U) contains X(99).

We now know that every nK which can be obtained by inversion with pole F is a cK(#F,U).
We might ask whether every cK(#F,U) is obtained by inversion. If we regard U as variable
and solve for P, we get a nasty cubic. This has obviously at least one real root, however,
so we have

Theorem 3
Every cK(#F,U) is the (P,F)-inverse of a circumconic.

However, even when F and U are triangle centres, P and K need not be - indeed for the
Tucker Nodal cubic cK(#G,G), we get P = [1,1,-5], K = [1,1,-2] (or rotations of these).

main tucker page