Several known cubics can be described as the loci of points for which the mid-point of X, iF(X)
lies on a fixed line. These include K040,K072,K086,K164,K165,K166,K185.

We can view the mid-point of PQ as the harmonic of the intersection of PQ with the tripolar
of G with respect to P,Q. This offers an easy, projective, generalisation.

Definition
Suppose that K is a fixed point. For points P,Q the K-harmonic of PQ is the harmonic of the
intersection of PQ and the tripolar of K with respect to P,Q.

Note that we can define this unless both P and Q lie on the tripolar of K. When P lies on the
tripolar, the K-harmonic of P and Q is P, whatever the choice of Q.

Thus, the G-harmonic of PQ is the mid-point unless one of P,Q is infinite.

If K = [k,l,m], P = [p,q,r], Q = [p',q',r'], then PQ has equation Σ(qr'-q'r)x = 0.
This meets the tripolar of K at [(p/k+q/l+r/m)p'-(p'/k+q'/l+r'/m)p]. Thus the harmonic is
[(p/k+q/l+r/m)p'+(p'/k+q'/l+r'/m)p].

We shall concentrate on the case where Q = iF(P) for the point F = [f,g,h]. In particular,
we look at K = G and K = F. The latter will generalise mid-point results and simson lines.

For the moment, we look at the general nK cubic :

        Σrx((hy)²+(gz)²) + Sxyz = 0.

with R = [r,s,t], F = [f,g,h], S symmetric.
For P = [x,y,z], the K-harmonic of P and iF(P) lies on the line L:ΣXx = 0 if and only if
Σ(Y/m+Z/l)x((gy)²+(hy)²) + 2(Σf²/k)xyz = 0.
Ths gives the points of the cubic if and only if there exists a real λ with
Yl+Zm = λrlm,
Zm+Xk = λskm,
Xk+Yl = λtkl,
2(ΣXf²/k) = λS.

The solution of the first three is given by
Xk = -r/k+s/l+t/m,
Yl = r/k-s/l+t/m,
Zm = r/k+s/l-t/m,
λ = 2/klm.
Then the fourth requires
(2)     Σ(-r/k+s/l+t/m)f²/k² = S/klm.

Note that, when the condition is satisfied for K = G, the equation of the line of
mid-points has coefficients X,Y,Z the coordinates of the anticomplement of R.

Note that if R lies on a fixed line αx+βy+γz = 0, then the line of mid-points
passes through the fixed point [β+γ].

We can clearly choose S to achieve any given line from any R,F.

The case of an nK0

For S = 0, there are many examples. In general the condition can be
written Σ(-(f/k)²+(g/l)²+(h/m)²)(r/k) = 0, i.e. the root lies on a line
related to F.

For K = G, the condition becomes Σ(-r+s+t)f²=0, or Σ(-f²+g²+h²)r = 0.
It follows that the line passes through [f²].
For F = I, we have isogonal conjugation. R lies on the orthic axis, and all
the mid-point lines pass through X(6). This is Gibert's class CL0025.
Gibert's list includes K018,K019,K040,K188,K189,K190.

For K = F, the condition becomes Σr/f = 0, i.e. the root lies on the tripolar
of F., or alternatively that F lies on C(R), the circumconic with perspector R.
The line is then Σ(r/f²)x = 0, the tripolar of iF(R) = the tangent to C(R) at F.
It contains the point F.

The case of a conico-pivotal conic

Here we have S = -2Σrgh. Then (2) becomes Σ(r/k)((g/l-h/m)²-(f/k)²) = 0.
This factorises to give the condition as Σr/k(-f/k+g/l+h/m) = 0.

Suppose that K = G.
Then the mid-point lines all pass through the G-Ceva conjugate of F
(as the complement of the isotome of the anticomplement of F).

Then, for F = I, we get mid-point lines through X(9) - a new class.
The root R lies on the tripolar of the Nagel point X(8).
The class includes K040 again! It is the only nK0 in the class.
One obvious point on the tripolar of X(8) is the infinite point X(522).
This gives as the line of mid-points X(2)X(9) = X(2)X(7).
The only other named point on the line is X(1639).

For K = F, the condition is equivalent to any of the following
1. the quotient R/F is at infinity,
2. the cubic is an nK0, (so we have the same as the case above),
3. R = P, the perspector of the associated conic,
4. R lies on the tripolar of F,
5. F is on C(R) the circumconic with perspector R
We also have that [1/f] lies on C*(F,R) so the tripolar of F
touches the pivotal conic.
The line is Σ(r/f²)x = 0, the tangent to C(R) at F. This contains F.

Other cases

Another obvious choice is S = -2Σrgh. so the cubic can be written
      Σrx(gz+hy)² = 0.
Then (2) becomes Σ(r/k)((g/l+h/m)²-(f/k)²) = 0.
This factorises to give (f/k+g/l+h/m) = 0, or Σ(r/k)(-f/k+g/l+h/m)= 0.

The former is that F is on the tripolar of K. It also lies on the circumconic K
with perspector iF(K), indeed the tripolar is the tangent to K at F.
This case may be degenerate.
If R = K, the cubic factorises as the tripolar of K and the circumconic K
This uses the condition f/k+g/l+h/m = 0. It amounts to the observation that,
for P on the tripolar of K, the K-harmonic of P and Q is always P.

The latter is that [r/k] lies on a certain line defined by F and K,
or [f/k] on a line defined by R and K.

For K = G, this means that R lies on the line Σ(-f+g+h)x = 0, or that
F lies on the line Σ(-r+s+t)x = 0 - the line of mid-points!

When F = I, we get Gibert's class CL003. This contains K040,K086,K165.
The mid-point lines all pass through I.

In general, we get a mid-point line through G by taking R at infinity.
One special case is when we take S = -2Σrf². Then the condition is
that R lie at infinity, so the line of mid-points is through G.

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