Suppose that U = u:v:w is a point at infinity.

Let **C** be the circumconic with centre U. This is a parabola.

Its perspector is U^{2}, the barycentric square of U.

This lies on the Steiner Inellipse.

Let **I** be the inconic with centre U. This is a parabola.

Its perspector is tU, the isotomic conjugate of U.

This lies on the Steiner Ellipse.

The focus F of **I** is gU, the isogonal conjugate of U.

This lies on the Circumcircle.

The axis **A** of **I** is then the line joining U and gU.

The directrix **D** of **I**, as the polar of the focus, is the tripolar

of the barycentric product of tU and H, the orthocentre. It now

follows that H lies on the directrix.

The axis of **C** has direction U, so is parallel to that of **I**.

Now for some less trivial observations.

**Observations 1**

(a) The polar of G in **I** is **L**, the tripolar of tU^{2}.

(b) **L** is the tangent to the Steiner Ellipse at tU.

(c) The intersections X,Y, of **C** and **I** other than U lie on **L**.

(d) **C** and **L** are isotomic conjugate.

(e) The points in (c) are the Mineur points of **C**.

*Proof.* Parts (a) and (b) are simple calculations. Part (c) is

proved by noting that the sum of their standard equations

factorises since U is at infinity. One factor gives **L**.

Parts (d) and (e) are now obvious.

**Observation 2**

The fourth intersection E of **C** with the Circumcircle lies on

the line **A**, the axis of **I**.

*Proof.* This is a straight-forward computation.

We now know that E and F are the intersections of **A** with the

Circumcircle.

**Observation 3**

The tangent to the Steiner Inellipse at U^{2} is the tripolar of U.

This is parallel to **L**, the common direction is given by the point

V = u(v-w):v(w-u):w(u-v).

*Proof.* Again, these are easy calculations.

**Observations 4**

(a) **I** and the tripolar of tU are tU-Ceva conjugate.

(b) The points X,Y in Observation 1(c) correspond to the

intersections of the tripolar of tU and the Steiner Ellipse.

*Proof.* Part (a) is a special case of the general result that the

inconic with perspector P and the tripolar of P are P-Ceva

conjugate.

Part (b) appears to need a Maple computation.

The point V in Observation 3 has an interesting property.

**Observation 5**

Let the lines AV, BV, CV meet **C** again in A', B', C'.

Then **I** touches the sidelines of **Δ'** = ΔA'B'C'.

*Proof.* This is a result from another piece of work, but it can

be verified by routine computation. We can find coordinates

for A', B', C' and simply check that, say, B'C' touches **I**.

Poncelet's Porism asserts that any point on **C** is a vertex of

a triangle inscribed in **C** and touching **I**.

**Observations 6**

(a) The triangle **Δ'** is the only member of this family which is

perspective with **Δ** = ΔABC.

(b) The centroids of **Δ** and **Δ'** coincide.

(c) The orthocentres of **Δ** and **Δ'** lie on a line parallel to the

directrix **D**.

(d) The circumcentres of **Δ** and **Δ'** lie on a line parallel to the

directrix **D**.

*Proof.* We intend to present a proof of (a) elsewhere, but it

can (nearly) be deduced from later observations.

For the rest, we note that several of the above objects are

defined by **C** and **I** without reference to **Δ**.

For example, the line XY is given by the intersections other

than U. The pole in **I** was identified as the centroid of **Δ**. It

must therefore be the centroid of **Δ'**. This is (b).

In the preamble, we observed that the orthocentre of a

triangle is on the directrix of any inparabola. This gives (c).

Finally, we noted that F, the focus of **I** and E, the intersection

(other than U) of **C** with the axis of **I**, lie on the Circumcircle

of **Δ**. Thus the perpendicular bisector of EF passes through the

circumcentre of **Δ**. Note that this bisector is then parallel to the

directrix. Then the circumcentre of **Δ'** is also on this line. This

is part (d).

As a digression, we give an observation which (almost) gives

Observation 6(a). From the rest of Observation 6, we see that

any poristic triangle perspective with **Δ** must have the same

centroid.

**Observation 7.1**

Suppose that W is any point not on **C**. Using W instead of V, we

define a triangle **Δ"**. Then the centroids of **Δ** and **Δ"** coincide

if and only if W = G or lies on a line. This has infinite point V.

*Proof.* This is a Maple computation using the coordinates of the

vertices of **Δ"**. The line is the dual of the complement of U^{2}.

To complete this proof, we need to discuss Fregier involutions.

Suppose that **E** is a conic, and P a point not on **E**.

For each point Q on **E**, we have a second intersection R of PQ

with **E**. This can be extended to an involution by the algebra,

or by geometry.

Suppose that we work in projective geometry with homogeneous

coordinates. The conic **E** has equation **x'Mx** = 0, where **x'**

denotes the transpose, and **M **is symmetric.

Then the Fregier map is given by F([**x**]) = [**x** -2(**p'Mx**/**p'Mp**)**p**].

Note that F is a projective map.

By routine algebra, F fixes Q =[**x**] if and only if Q = P, or Q is on

the polar of P in **E**.

Also, as F(Q) is on PQ, F maps ΔABC to a perspective triangle F(ΔABC).

If **E** is a circumconic of ΔABC, then it is a circumconic of F(ΔABC).

Likewise for inconics.

In our case, we have a triangle in the poristic family of **C** and **I**

which is perspective with ΔABC. Let F be the Fregier map defined

by **C** and the perspector P ot the triangles. The conic **I** is an

inconic of both triangles (we are in the poristic family). There is

only one conic touching even five of the six sidelines, so F fixes **I**.

As F maps Q to a point of PQ, it is also the Fregier map of P and **I**.

Here F fixes the point which is the centroid of both triangles, so 'P'

must be on both polars. Hence P = V. This gives Observation 6(a).

**Observation 8**

Let c(**I**) denote the complement of the inconic **I**.
Then

(a) c(**I**) is a parabola with centre U.

(b) c(**I**) is an inconic for the medial triangle of ΔABC.

(c) c(**I**) is a diagonal conic with respect to ΔABC.

*Proof.*

Parts (a) and (b) are trivial since complementation preserves U

and the line at infinity.

Part (c) is an easy calculation. The equation of c(**I**) is

x^{2}/u+y^{2}/v+z^{2}/w = 0.

It is easy to verify that, if **E** is a parabola with centre U, then, for

any P, the line PU cuts **E** at U and the mid-point of PQ, where Q is

the meet of PU and the polar of P in **E**. This is a simple case of a

result on harmonics and projective conics.

**Observation 9**

Suppose that

U3 is the barycentric cube of U. This lies on GU.

Q1,P1 are the intersections of GU with **C** and the polar of G in **C**.

Q2,P2 are the intersections of GU with **I** and the polar of G in **I**.

Q3,P3 are the intersections of GU with c(**I**) and the polar of G in c(**I**).

Then

The points P1,U3,Q1,P3,G,Q2,P2 are equally spaced along GU, and

Q3 is the mid-point of GP3.

*Proof.* These are easily checked by Maple since each point on GU3

is of the form kU3 + lG - i.e. has coordinates [ku^{3}+luvw].