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 non-standard:
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 G-Ceva 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).
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 K ellipse.
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
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
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 mid-point of X(1641) and X(1648) is X(2).
L touches the Steiner inellipse at X(115), the K ellipse at X(125).
Concentric circumconic and inconic.
The centre and perspector are G-Ceva conjugates, so we might expect geometrical relations.
The following results can be verified algebraically, but the result is purely geometric.
Suppose U = u:v:w, so that, in an obvious notation, G(U) = [u(v+w-u)].
1. The circumconics C, D with centres and perspectors U and G(U) meet at P = [u/(v-w)].
2. The mid-point of U and G(U) is Q = [u(uv+uw+2vw-u2)].
3. The centroid X(2) lies on PQ.
( Let E, F be the inconics with centres U and G(U).
3. The perspectors are [1/(v+w-u)] and [v+w-u], respectively
4. The inconics E, F meet at R = [(v+w-u)(v-w)2].
5. The point R lies on PQ.
U = X(1), so that G(U) = X(9).
Then P = X(100), Q = X(1001), R = X(11).
The inconics are the Mandart ellipse and the Incircle.
U = X(3), so that G(U) = X(6). Then P = X(110), Q = X(182), R = X(125).
The conics include the Circumcircle and the K ellipse.