Circumconics and inconics

Suppose we have a triangle T = ΔABC. We say that C is a circumconic for T if it passes through A,B and C.
We say that I is an inconic if it touches AB, BC and CA.
We take T* as the reference triangle with vertices X=[1,0,0],Y=[0,1,0], Z= [0,0,1].

It is easy to see that a circumconic for T* has an equation of the form pyz+qzx+rxy=0, which is non-degenerate
provided that pqr is non-zero.

If AA',BB' and CC' meet in a point P, then we say that the triangles ΔABC and ΔA'B'C' are in perspective from P.
We refer to P as the perspector of the triangles.

Desargue's Theorem
If triangles T = ΔABC and T' = ΔA'B'C' are in perspective from a point, then the intersections of AB,A'B', of BC,B'C'
and of CA,C'A' are collinear.

We refer to the line of intersections as the perspectrix of the triangles.

Theorem 1
If C is a circumconic of triangle T = ΔABC, then the tangents to C at A,B,C from a triangle in perpsective with T.
If T = T* and C: pyz+qzx+rxy=0, then the perspector is [p,q,r], and the perspectrix is x/p+y/q+r/z=0.

We give a proof later.

In general, we refer to the perspector of T and its tangential triangle as the C(T)-perspector. For T*, we refer
to it simply as the C-perspector.

We often use duality with respect to the fixed conic D: x2+y2+z2 = 0. Although this has no real points, it is
non-degenerate as a complex conic.
Duality with respect to D associates the point [u,v,w] with the line ux+vy+wz=0. Note that if the point is real,
so is the line. Also, the reference points X,Y,Z correspond to the lines x=0, y=0, z=0, the reference
triangle T* is self-dual.
Since D has matrix I, earlier results show that, if a conic C has matrix M, then its D-dual has matrix M-1.
Again, if C is real, so is its dual.

As a first application we note that the D-dual of an inconic of T is a circumconic of the dual triangle, and the
contact triangle for T corresponds to the tangential triangle for the dual. Theorem 1 shows that the latter
pair are in perspective. Duality shows that the perspectrix corresponds to a perspector for T and its contact
triangle. Thus we have

Theorem 2
If triangle T has inconic I, then T and the triangle with vertices at the contacts of T with C are in perspective.

As before, we talk of the I(T)- and I-perspectors. We can get results for coordinates from the rest of Theorem 1.
The reference triangle T* is self-dual, so if the inconic I corresponds to the circumconic C : pyz+qzx+rxy = 0,
then the I-perspector is [1/p,1/q,1/r]. Using the matrix for C, we get an equation for I namely

I : p2x2+q2y2+r2z2-2pqxy-2qryz-2rpxz = 0.

We summarise some results on poles and polars of conics as follows :
Suppose that a conic C has matrix M. Then
(1) the polar of the point U= [u] = [u,v,w] is the line xTMu = 0,
(2) the pole of the line L : ux+vy+wz = 0 is the point M-1u.

If C is the circumconic pyz+qzx+rxy = 0 and U = [u,v,w], then (1) allows us to write the tangent to C at U as

U*: px/u2+qy/v2+rz/w2 = 0,
provided uvw is non-zero. Of course the exceptions are simply the reference points X,Y,Z. For these, (1) gives
the tangents directly as
X* : ry+qz = 0,
Y* : rx+pz = 0,
Z* : qx+py = 0.

Proof of Theorem 1
For the reference triangle and the given conic, the tangents Y*, Z* meet at X** = [-p,q,r].
The line XX** has equation ry-qz = 0. This passes through P = [p,q,r]. By algebraic symmetry, P is also on
the lines YY** and ZZ**. The general case follows since we can apply a projective transformation to map the
vertices of any triangle to X,Y,Z.

inconics and the newton line