This represents a new approach to the cK(#F,R) and offers a simple construction.
We begin with a fairly obvious map from a line to a cubic.
Let K = cK(#F,R), with F = [f,g,h], R = [r,s,t].
For U = [u,v,w], write L(U) for the dual of U, the line ux+vy+wz = 0.
Note that U is on L(F) if and only if F is on L(U) as each means fu+gv+hw = 0.
As F is the node of K, any line through F meets K twice at F, so there is a third
intersection of the line with K.
Theorem 1
If U = [u,v,w] is on L(F), then the third intersection of L(U) and K is the point
dt(U) = [(sv-tw)/u,(tv-ru)/s,((ru-sv)/t].
This was obtained by solving equations for L(U) and K, but is trivial to verify.
The line L(U) will therefore be tangent to K at F - a nodal tangent - if and only if
the "third" intersection is also F.
Theorem 2
If U = [u,v,w] is on L(F), then L(U) is a nodal tangent if and only if U lies on the
conic C1 : rfx2+sfy2+thz2 = 0.
Proof
The condition that L(U) is a nodal tangent is that dt(U) = F, i.e. [u,v,w]=[nf,ng,nh]
for some non-zero n. From the first two coordinates, we get the condition that
gv(sv-tw)-fu(tw-ru) = 0, i.e. sgv2+rfu2-tw(gv+fu) = 0.
But P is on L(F), so -(gv+fu)=hw, and the condition is that P is on C1.
This is symmetric, so it is also the condition that the first and third coordinates
agree, i.e. that P = F.
It follows that the nodal tangents are L(U), where U is on C1 and L(F). Then we have
U is on C1 and L(F) if and only if L(U) - the dual of U - is tangent to the dual of C1
and passes through F - the dual of F. We have proved
Corollary 2.1
The nodal tangents of K are the tangents from F to the conic C2 : x2/rf+y2/sg+z2/th = 0.
Observe also that the contact points of the nodal tangents with C2 will lie on the polar of F
with respect to C2. This is x/r+y/s+z/t = 0 - the tripolar of R.
Corollary 2.2
The nodal tangent pair for K has equation
x2(gt+hs)/f+y2(hr+ft)/g+z2(fs+gr)/h-2txy-2ryz-2szx= 0.
This is a consequence of a general result for the tangent pair from a point to a conic.
If the conic is x'Mx = 0, the tangent pair from u is (u'Mu)(x'Mx) = (x'Mu)2.
The intersections of C2 and the tripolar of R are easily found. If we eliminate x from the
equations, we get a quadratic with discriminant Δ = -(rs/fg+st/gh+tr/hf)/(st)2.
When K = cK0(#F,R) - so r/f+s/g+t/h = 0 - Δ is positive so we have real nodal tangents
and real conics C1, C2.
Degenerate cases
The nodal tangents coincide if and only if Δ = 0. But this is precisely the case when K
degenerates - as the tripolar of R and the conic with perspector the (F2-)isoconjugate of R.
This is easy to show. If the cubic degenerates, one factor must be the tripolar of R since
K
contains the intersections of this line with the sidelines. The remaining conic must then
have the given perspector to get the correct terms other than that in xyz. Finally, we get
the correct xyz term precisely when Δ = 0.
Isoconjugates
As K is an isocubic, if X = [x,y,z] is on K, then so is its (F2-)isoconjugate [f2/x,g2/y,h2/z].
We can actually identify the point which gives rise to the isocongugate of dt(U).
Theorem 3
If U = [u,v,w] is on L(F), then dt(U)* = dt(V), where
V = di(U) = [(sv-tw)/f,(tw-ru)/g,(ru-sv)/h)].
Proof
It is simple to verify that, for this V, dt(V) = [f2u/(sv-tw),g2v/(tw-ru),h2w/(ru-sg)] = dt(U)*.
This is related to the conics C1 and C2:
Theorem 4
(1) If U = [u,v,w] is on L(F), then the pole of L(P) with respect to C2 lies on L(di(U)).
(2) If U = [u,v,w] is on L(F), then the polar of U with respect to C1 passes through di(U).
Proof
(1) The pole is [fru,gsv,htw], so lies on the required line.
(2) The polar is frux+gsvy+htwz = 0, which contains the given point.
Of course, these are dual results.
The map dt is from L(F) to K. It takes U on L(F) to the third intersection of L(U) and K.
The inverse map takes X on K to the intersection of L(F) and L(X), since X is on L(U)
if and only if U is on L(X), and we require U on L(F). Thus we have
Theorem 5
If X = [x,y,z] is on K then X = dt(U) where U = [hy-gz,fz-hx,gx-fy].
U can be constructed as the intersection of L(X) and L(F).
An alternative description
Here, we obtain a map from C(F), the circumconic with perspector F, to K. This turns out to
have nicer properties. Observe first that U is on C(F) if and only if its isotomic conjugate U'
is on L(F). Also, the dual of U' is the tripolar of U.
Applying Theorem 2 to the isotomic conjugate, we get
Theorem 6
If U = [u,v,w] is on C(F), then the third intersection of the tripolar of U and K is the point
tt(U) = [u(s/v-t/w),v(t/w-r/u),z(r/u-s/v)].
tt(U) is the intersection of the tripolar of U and the tripolar of the U2-isoconjugate of R.
The last part is a simple observation once we have the coordinates of tt(U). It allows us to
construct K quickly from C(F) and R.
We get an analogue of Theorem 5 by taking the isotomic conjugate. Here, the point V has
a nice description using the formula in Theorem 5.
Theorem 7
If X is on K(F), then X = tt(V), where V is the tripole of XF.
There is also an analogue of Theorem 3. Once again, the geometry is simple.
Theorem 8
If U = [u,v,w] is on C(F), then tt(U)* = tt(V), with V the second intersection of C(F) and UR.
Proof
We know that tt(U) = dt(U'), where U' =[u',v',w'], the isotome of U, lies on L(F).
By Theorem 3, dt(U')* = dt(W), where W = di(U'). Let V = W'. From Theorem 3,
V = [f/(sv'-tw'),g/(tw'-ru'),h/(ru'-sv')]. Now U = [1/u',1/v',1/w'], so that UR has
equation u'(sv'-tw')x+v'(tw'-ru')y+w'(ru'-sv')z = 0. V is on UR as U' is on L(F).
Of course, there are degenerate cases when UR is tangent to C(F). For such U,
tt(U) must be F as a the only point of K fixed by the isoconjugation. We now have
Theorem 9
The nodal tangents are the tripolars of the intersections of C(F) and the polar of R.
Proof
In Theorem 2, we saw that the nodal tangents were the duals of the points on L(F)
which give rise to F under dt. It follows that they are the tripolars of those points
of C(F) which give rise to F. As noted above, these are the points of contact of C(F)
with the tangents from R. These are the intersections of C(F) and the polar of R.
When K is a cK0, the polar in Theorem 9 can be described in other ways.
Corollary 9.1
If K = cK0(#F,R), then the nodal tangents are the tripolars of the intersections of C(F)
with the line L which is
(a) the polar of R with respect to C(F),
(b) the tripolar of the isoconjugate of R,
(c) the tangent to C(R) at F.
Using other conics
We chose to work with the conic C(F) since this leads to a map to K such that the domain
depends only on F, and the rule only on R.
There are advantages in other choices of conic. Let us choose C(L), where L = [l,m,n].
Each point U" on C(L) has the form [lu/f,mv/g,hw/h] with U = [u,v,w] on C(F). We can
associate with U" the point tt(U) on K. This map then takes X = (x,y,z) to the point
t"(L,X) = [(fx/l)(ms/gy-nt/hz),(gy/m)(nt/hz-lr/fx),(hz/n)(lr/fx-ms/gy)].
Theorem 10
The map t(L,F) above has the property that t(L,V) = t(L,U)* if and only if
V is the second intersection of C(L) and UR", where R" = [lr/f,ms/g,nt/h].
If we choose L = the complement of the barycentric quotient R/F, then
t(L,V) = t(L,U)* if and only if U,V are antipodes on C(L).
Proof
In the proof of Theorem 8, we observed that tt(V) = tt(U)* precisely when R lay on UV.
It follows that t"(V") = t"(U")* precisely when R" lies on U"V".
When L is the stated complement, R/F is the anticomplement of L, so the centre of C(L)
is the product L(R/F). This is R".
Note that this is a generalization of Gibert's work on the Simson Cubic. We have not,
however, identified t(L,X) in any geometric way - e.g as the tripole of a significant line.