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 U2, 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.
(a) The polar of G in I is L, the tripolar of tU2.
(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.
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
The tangent to the Steiner Inellipse at U2 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.
(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
Part (b) appears to need a Maple computation.
The point V in Observation 3 has an interesting property.
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.
(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
(d) The circumcentres of Δ and Δ' lie on a line parallel to the
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
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 U2.
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).
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.
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
x2/u+y2/v+z2/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.
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).
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 [ku3+luvw].