These are some random notes on circumconics and inconics and duality.
We
work exclusively in barycentrics. We use the following notation, some
of
which is nonstandard:
CP(U )is the circumconic with perspector U,
CC(U) is the circumconic with
centre U.
IP(U )is the inconic with perspector U,
IC(U) is the inconic
with centre U.
G(U) is the GCeva conjugate of U.
C(U) is the complement
of U.
A(U) is the anticomplement of U.
TU) is the isotomic conjugate of
U.
[f(a,b,c)] is the point with barycentrics f(a,b,c):f(b,c,a):f(c,a,b).
Thus,
G([f(a,b,c)]) =
[f(a,b,c).(f(b,c,a)+f(b,c,a)f(a,b,c))].
C([f(a,b,c)]) =
[f(b,c,a)+f(b,c,a)].
A([f(a,b,c)]) =
[f(b,c,a)+f(b,c,a)f(a,b,c)].
T([f(a,b,c)]) = [1/f(a,b,c)].
We have some familiar results:
If a circumconic has perspector U, then it
has centre G(U).
If a circumconic has centre U, then it has perspector
G(U).
If a inconic has perspector U, then it has centre C(T(U)).
If a
inconic has centre U, then it has perspector T(A(U)).
Suppose that U = u:v:w is a point, then
ux+vy+wz = 0 is the dual of
U.
x/u+y/v+z/w = 0 is the tripolar of U.
Suppose that L : fx+gy+hz
= 0 is a line, then
f:g:h is the dual of L.
1/f:1/g:1/h is the
tripole of L.
The duals of the points of a conic C envelop a conic  the dual C' of
C.
The duals of the tangents to a conic D constitute the dual D' of D.
Then
for any conic E, E'' = E.
If the circumconic C has perspector U, then its dual C' is the inconic with
perspector T(U) and vice versa.
If U = u:v:w, then CP(U) has equation u/x+v/y+w/z = 0, so that CP(U) is
the isotomic conjugate ot the dual of U. Combining this with earlier
results,
we see that the dual of IP(V) is the isotomic conjugate of the
tripolar of V.
We use Clark Kimberling's notation for triangle centres. Thus, the
incentre
is X(1), the centroid is X(2), and so on.

 







 







 







 







 







 







 







 







 







 







 









 
















































































































a remarkable coincidence
Of the eleven named inconics, five have duals through X(99).
By duality,
the inconics have a common tangent L, the dual of X(99).
This is the line
through X(115) and X(125). The line is the tripolar of X(523).
We may add to the list the dual of the circumcircle as this contains X(99).
A result going back to Newton shows that the inconics touching L lie on
a
line L* known as the Newton line for L. In general, if L has
equation
fx+gy+hz = 0, then L* has equation Fx+Gy+Hz = 0, where [F] =
A(T([f])).
Here the situation is particularly simple. When [f] is on the Steiner
ellipse,
so T([f]) is at infinity, and then [F] = T([f]). Here, [f] = X(99),
so the Newton
line is the tripolar of X(99). This is the line though X(2) and
X(6). This comes
directly from the fact that the list includes the Steiner
inellipse and the Orthic inconic.
We can also locate the perspectors of these inconics. Since the dual
circumconics
pass through X(99), the perspectors of the circumconics must lie
on the tripolar
of X(99). The perspectors of the inconics are the isotomic
conjugates of these
must lie on the isotomic conjugate of this tripolar. Thus
the perspectors lie on
the circumconic with perspector X(523], the Kiepert
hyperbola.
The line L* has 61 named centres, so we could extend our list of inconics
which
touch L.
The family of inconics includes one parabola P. This must have centre at
infinity
and perspector on the Steiner ellipse. Thus, P has centre X(524) and
perspector
X(671).
It must also contain the inconic with L as asymptote. By other results, this
has as
its centre X(1648), the tripolar centroid of X(523). We could also see
this as the
intersection of L and L*. The list ETC shows that X(1641)  the
tripolar centroid
of X(99)  lies on L*, so is the centre of one of the
inconics.
Note that the midpoint of X(1641) and X(1648) is X(2).
L touches the Steiner inellipse at X(115), the Orthic inconic at X(125).
Concentric circumconics and inconics.
Suppose that U = u:v:v and V are isotomic conjugates, so that V =vw:wu:uv.
The inconics with perspectors U, V have centres C(V), C(U) respectively.
Let G be the centroid of ΔABC. Then C(U),C(V) are GCeva conjugate.
The midpoint of C(U)C(V) is M = [(u^{2}+vw+2uv+2uw)(v+w)]
The line GM has equation fx+gy+hz = 0, where f:g:h = [(vw)(u^{2}vw)].
The circumconics with centres C(V),C(U) have perspectors C(U),C(V),
so meet at P = [(v+w)/(vw)], as well as at A,B,C.
The inconics with centres C(V),C(U) have perspectors U,V,
so meet at Q = [u(vw)^{2}], and at three other points as well.
Maple calculation verifies that P and Q lie on GM,
and that Q is the complement of P. It follows that
the complement of Q is the midpoint of PQ.
Thus, Q lies on the complements of the circumconics.
Each such complement is a circumconic of the medial triangle.
The inconic with perspector U touches BC at 0:v:w  the
intersection of AU and BC. This is the complement of the point
v+w:wv:vw. This lies on the circumconic with perspector C(V),
i.e. with centre C(U). Thus, the complement of the circumconic
with centre C(U) passes through the vertices of the UCevian
triangle. Similarly, the complement of the circumconic with
centre C(V) passes through the vertices of the VCevian triangle.
Since complementation is an affine transformation, it preserves
centres and perspectors. Thus the complements of the conics have
a common {centre,perspector} pair. We can then define inconics
for the medial triangle, and so on ..........
In fact, we can say more about the above inconics.
They contain the points [u(v^{2}w^{2})^{2}] and [u^{3}(v^{2}w^{2})^{2}].
These are the contact points with the fourth common tangent, which is the
tripolar of [u(v^{2}w^{2})] (by duality).
This is part of a general result:
If U = u:v:w and P = p:q:r are general points, then the inconics with perspectors
U and P have as their fourth common tangent the tripolar of [up(vrwq)].
The contact points are [up^{2}(vrqw)^{2}] and [u^{2}p(vrwq)^{2}], respectively.
For general V, the line GN where N is the midpoint of U,V is itself quite interesting.
Suppose that f,g are symmetric, homogeneous and of the same degree in a,b,c.
For U = u:v:w, let U(f,g) = [fu+g(v+w)]. Then U(1,0) = U, U(0,1)=C(U),
U(1,1) = A(U), U(1,1) = G. Also, for any f,g U(f,g) lies on GU and we have
UU(f,g)/U(f,g)U(0,1) = g/f.
For any f,g, the midpoint of U(f,g) and V(f,g) lies on GN.
If h,k is another such pair, then
the intersection of U(f,g)V(h,k) and U(h,k)V(f,g) lies on GN.
We have the special case that the midpoint of A(U)and A(V) is the
intersection of UC(V) and C(U)V.
examples of conics
U = X(7), V = X(8).
C(U) = X(9), C(V) = X(1).
M = X(1001), P = X(100), Q = X(11).
The inconics are the Incircle and the Mandart ellipse.
U = X(69), V = X(4).
C(U) = X(6), C(V) = X(3).
M = X(182), P = X(110), Q = X(125).
The circumconics are the Circumcircle and the dual of the Macbeath inconic.
The second inconic is the Orthic inconic.
Both examples have P on the Circumcircle and Q on the ninepoint circle.
This is a coincidence. By looking at the equations λf = (v+w)/(vw),
λg = (w+u)/(wu), λh = (u+v)/(uv), we see that P is only constrained
to lie outside the Steiner ellipse. Then its complement Q is outside the
Steiner inellipse.