Hirst-type cubics and the generalised Simson map

Note We should note that we are actually discussing the class cK0.

We use the notation of the Simson and Hirst pages.

We have a cubic K = cK(#F,R) of Hirst type, so that R/F is at infinity.
We have two maps from the associated conic R to K.

As always, we have the generalised simson transform, which we can write as
           gs : [x] -> [r(x/u)(v/y-w/z)],
where U is the cross-point of R and F.
This is undefined for X = U. In general gs(X) = g(X') if and only if X' = X or
X' = [x/(-u/x+v/y+w/z)], the U-cross conjugate of X.
It has no fixed points. It maps F to F1 = [f(t/h-s/g)], on the tripolar of F.
Of course, the isoconjugate F2 = [f/(t/h-s/g)] will also be a point of K.
As F1 is on the tripolar of F, F2 is on the circumconic F. It is trivial to
check that it is the fourth intersection of F and R.
As F is on R in these cases, this gives an intersection of K and the tripolar.
It is easy to check that F' = R and gs(R) = gs(F).
Maple also shows that there are two points which map to F. They are also points for
which X = X', with X' as above.
They are the points where R meets the quadric Q (see the simson page).
It follows from this that they lie on R. Let these points be Q, Q". As F isoconjugates
to itself, QQ" must pass through the point U, (see simson, Theorem 3).

We can actually prove much of this without Maple. We know that gs maps both
R and Q to K. If we take Q as one point on R which maps to F, then the U-antipode
maps to the isoconjugate of F, i.e. to F again. Thus both lie on R. The map X -> X'
above must interchange R and Q, so Q, Q' lie on Q.

If we use the geometry of gs, we can construct a partial inverse.
Suppose X is a point. Let L be the tripolar of X.
Let A', B', C' be the intersections of L with the sidelines.
Let A", B" C" be the intersections of the tripolar of F with the cevians of F.
Then we can write down the equations of A'A", B'B", C'C".
Maple shows that these are concurrent if and only if X lies on the cubic K.
Since this process reverses the construction of the simson transform, we
have an inverse for gs restricted to a map R to K.
The results also offer a formula. To make this symmetric, we can take the
points of intersections of the above lines in pairs. Then we add respective
coordinates to get a result which is a centre. The formulae are unpleasant.
Each coordinate is quadratic in the coordinates of X.

Note that we have identified two points on K with coordinates purely in terms
of the coordinates of R and F. the circumconic can be identified by F and the
second of these points. These points can be defined even if K is not of Hirst
type, though the coordinates are slightly more complicated. the first is
[f2((t/h)2-(s/g)2)/r]. The second is its isoconjugate.

Since K is of Hirst type, we also have F-Hirst inversion
           hf : [x] -> [ghx2-f2yz].
Algebraically, hf is undefined for X = [f,θg,θ2h], θ3 = 1. Let these be F, F', F".
Note that F' and F" are isoconjugates.
Also, F' and F" lie on F, so, geometrically, hf fixes F' and F". F, F', F" lie on K.
For X on the tripolar of F, hf(X) = F.
For X ≠ F on the tripolar of F', hf(X) =F", and similarly for F".
For points not on these tripolars, hf is an inversion. It fixes the the points of F.
This is obvious from the geometry - X lies on its polar if and only if X is on F.

In hirst, we saw that hf maps R to K. It maps U-antipodes on R to isoconjugates
on K.

Checking by Maple, we find that:

K and F meet in A,B,C,F',F" and F2 = [ 1/(vh-wg)].
F and the tripolar of F meet in F',F".
K and the tripolar of F meet in F',F",F1 = [f2(vh-wg)].
K and R meet in A,B,C,F2 and F twice.
The tripolars of F and the isoconjugate of R meet at F1.
F and R meet in A,B,C,F2.
F2 is the U-antipode of F since U = [r2/f], so F2 is on UF.
(F',F") and (F1,F2) are isoconjugate pairs.
F2 is the cross-point of F and F1.

Note that the line of F-harmonics is FF1, which is the tripolar of the isoconjugate
of R, and also the tangent to R at F.

The points F', F" must be images of points on R. They are images of points on
their tripolars. It is easy to check that these are the points R' = [r,sθ,tθ2] and
R" = [r,sθ2,tθ] - the intersections of R with the tripolar of R. U is on R'R", as
expected, since F',F" are isoconjugate. Note that the tripolars of F',F" meet at F,
so we have found the only other intersections with R.

Note that the tripolar of F is also the polar of U with respect to R. The tripolar of F
meets R in two points T1,T2 which hf maps to F. These are their own U-antipodes.
This proves the first part of the next result.

Theorem 1
The tangents from U to R are UT1, UT2.
The nodal tangents of K are FT1, FT2.

Proof
A typical line L through F has the form kx/f+ly/g+mz/h = 0, with k+l+m = 0.
This meets K at F and one other point S. L meets the tripolar of F at the point
T = [f(l-m)]. Maple shows that the condition for S = F is the same as that for
T to be on R. T is on the tripolar of F, so T = T1 or T2.

Note that the nodal tangents are also tangent to the pivotal conic, so we have the

Corollary The pivotal conic is inscribed in ΔFT1T2.

Of course, the pivotal conic is also inscribed in the anticevian triangles of F,R.

The calculation for the above Theorem actually yields more.

Theorem 2
For a cubic cK(#F,R), the nodal tangents are the tangents from F
to the conic D = Σ(1/rf)x2 = 0.

When the cubic is of Hirst type,
D is the circumconic to the anticevian triangle of R which passes through R
with the tripolar of F as the tangent at R.
D is the conic inscribed in the cevian triangle of F, tangent to the tripolar
of F at R.
Its centre is R*F (barycentric). In either case the perspector is U as above.

Proof
Let L be the line Σkx/f = 0 - note that this is the dual of [k/f].
The condition for L to be a nodal tangent is Σrk2/f = 0.
Now observe that the required [k/f] are the solutions of Σrfx2 = 0 and Σfx = 0.
Thus the [k/f] are the intersections of a conic with a line. Taking duals, we see
that the tangents are the tangents to the dual conic D = Σ(1/rf)x2 = 0 from F.
The observations on D are trivial.

Note: Maple worksheet nodaltangents.mws

Footnote

We have two maps from R to K. Each maps U-antipodes to isoconjugate points.
Our work above shows that the maps are different since, for example, the simson
map has no fixed points. They different even on R, the hirst map fixes F2.

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