Inconics - part 3 : two views of an inconic

We have described an inconic in two different ways - as the Ceva conjugate of a tripolar
and as the locus of generalised orthopoles of lines through a fixed point. The former is
easily seen to be equivalent to the description as the dual of a circumconic. Here, we will
look at the relationship of the two views.

For the first, see inconics, part 1, and inconics, part 2.
For the second, see generalised orthopoles, and inconics and desmic structures.

Recall our notation :

C(X) is the circumconic with perspector X,
I(X) is the inconic with perspector X,
T(X) is the tripolar of X,
D is any point,
K = cD, the complement of D,
H = tD, the isotomic conjugate of D,
L = aH, the anticomplement of L,
O = cH, the centre of the circumconic C(K).,
G denotes the usual centroid.

Basic Result
The inconic I(D) is
(a) the locus of D-Ceva conjugates of points on T(D), and
(b) the locus of K-orthopoles of lines through L.

The first description is easy to implement. Each point X on T(D) gives rise to a unique
point on I(D). It is the contact of I(D) with T(X), which is a tangent to I(D). For the other,
we have to select lines through L. To make the algebra easier, we can classify the lines
by their points at infinity. This does not appear to have much geometric significance.

To describe the link, we introduce a further point Uo. Suppose that D = p:q:r, Uo = u:v:w.
We shall also need the harmonics of Uo, these are Ua = -u:v:w, Ub = u:-v:w, Uc = u:v:-w.
Let D1 be the isotomic conjugate of the Uo2-isoconjugate of D, so D1 =p/u2:q/v2:r/w2.
Let D2 be the point q-r:r-p:p-q. This is the dual of the line GK.
For x = o,a,b,c, let Dx be the D-isoconjugate of Ux, e.g. Do = p/u:q/v:r/w.
We then have

Theorem 1
Let Mx be the line joining Ux and L (x = o,a,b,c).
The K-orthopoles of Mo, Ma, Mb, Mc are on I(D) and I(D1) if and only if
Uo (and hence each Ux) is on the diagonal conic D = (q-r)x2+(r-p)y2+(p-q)z2 = 0.
For such Ux, these orthopoles are the D-Ceva conjugates of the intersections of the
line T(D) with each of the lines T(Dx).

An important difference
In Inconics Part 2, we discussed the general intersection of inconics. These occur as the
D-Ceva conjugates of intersections with a tripolar and its harmonics. Here, we look for
cases where the intersections arise from generalised orthopoles defined by a point and
its harmonics. From the Theorem, we see that this only occurs for a restricted class of
inconics. From the detail of Theorem 1, we see that the inconics I(D1) must have their
perspectors on the circumconic p(q-r)yz+q(r-p)zx+r(p-q)zy = 0. This is the isotomic
conjugate of the line GH, the analogue of the Euler Line.

Notes on the conic D.
This contains G and its harmonics, the vertices of the antimedial triangle. It contains L.
Observe that these five points define the conic. Since L is on the conic, a line through
L meets D in one other point. This means that we have a mapping from D to I(D).
The conic also contains the four points which are fixed by K-isoconjugation, but these
may not be real.
The conic D is the anticomplement of the circumconic K = C(D2). This is the analogue
of the Kiepert Hyperbola. It contains G and H, so is the K-rectangular hyperbola whose
centre is on the Steiner Inellipse.

Recovering U from a point on I(D).
Given a point on D, the above theory shows how to construct the corresponding point
on I(D). We now consider the problem of recovering the point U on D which gives rise
to a given point on I(D).

Theorem 2
Suppose that X is a point on I(D). Then X is the K-orthopole of UL, with U on D where
U is given by the following construction.
Let Y be the D-Ceva conjugate of X.
Let Z be the barycentric product of D and the isotomic conjugate of Y.
Then U is the intersection of T(Z) and D other than G.

Proof notes
The proof is algebraic. We know that, if U = u:v:w, then the corresponding X has
coordinates p(v-w)2:q(w-u)2:r(u-v)2. Then Y is p(v-w):q(w-u):r(u-v). From this,
T(Z) has equation (v-w)x+(w-u)y+(u-v)z = 0. This contains G and U, so that these
are the intersections with D. Thus, U is as claimed.

Constructional note
This appears to require the intersection of a line and conic. We know enough points on
the conic D to construct U by the Breckenridge-Maclaurin method. We take the points L
G, and the vertices of the antimedial triangle on D. The method gives the other meeting
of D with the line T(Z) through G.

Two descriptions of the points of I(D)

The initial description is that each point of I(D) is the K-orthopole of a lines through L and
a point of D. It is rather difficult to give a neat algebraic characterization of the points of
the diagonal conic D. We can also say that each point of I(D) is the K-orthopole of the
anticomplement of a line through H and a point of the circumconic K. The points of this
are easy to describe since the x-coordinate is a rational function of the others.

Theorem 3
Let X = x:y:z be any point. Define the further points Y and Z by
Y = (q-r)(-x/p+y/q+z/r):(r-p)(x/p-y/q+z/r):(p-q)(x/p+y/q-z/r),
Z = (p-q)y/q+(r-p)z/r:(p-q)x/p+(q-r)z/r:(r-p)x/p+(q-r)y/q.
(1) Points U,V on K give points on I(D) collinear with X if and only if U,V,Y are collinear.
(2) Points U,V on D give points on I(D) collinear with X if and only if U,V,Z are collinear.

Proof notes
From the proof notes for Theorem 2, we know the coordinates of the point of I(D) derived
from the point U = u:v:w on D. The complement cU leads to the same point on I(D) from
the geometry, or from the algebra. A little thought shows that the point u:v:w on K leads
to the point on I(D) given by the formula above.
Using Maple, we can find conditions that
(a) two points U,V on K are collinear with an arbitrary point W, and
(b) U,V give rise to points on I(D) collinear with the point X.
We can then find the condition on W which makes these equivalent.
This gives the coordinates of W as linear in terms of those of X.
It turns out to be the point Y. Taking anticomplements, we get the condition on points of D
giving points on I(D) collinear with X. It is the condition that the points on D are collinear
with the point Z, the anticomplement of Y.

Some examples

(1) antipodes on I(D)
The centre of I(D) is p(q+r):q(r+p):r(p+q). Theorem 3 then gives
U,V on K (or D) give antipodes on I(D) if and only if U,V are collinear with the point
"X(525)" = p(q-r):q(r-p):r(p-q), the direction of T(D).

(2) D-antipodes on I(D)
Two points on I(D) are D-antipodes if they are collinear with the perspector D.
U,V on K (or D) give D-antipodes on I(D) if and only if U,V are collinear with the point
"X(523)" = (q-r):(r-p):(p-q), the direction of T(H).

(3) "X(525)"-antipodes on I(D)
U,V on K give "X(525)"-antipodes on I(D) if and only if U,V are collinear with the point
"X(115)" = (q-r)2:(r-p)2:(p-q)2, the centre of K.
U,V on D give "X(525)"-antipodes on I(D) if and only if U,V are collinear with the point
"X(99)" = 1/(q-r):1/(r-p):1/(p-q), the centre of D.

Note on Example 1
We know that lines through L give antipodal points precisely when they are perpendicular.
Example 1 relates points on K giving antipodal points on I(D). We shall discuss the case
where K = X(6) in detail. The general version is an easy consequence.
Points U,V on the Kiepert Hyperbola with UV in the direction X(525) give antipodal points
on I(D). Thus the lines UH, VH are perpendicular.
X(525) is the orthopoint of X(1503), the direction of the line KH.
The line KH is the tangent to the Kiepert Hyperbola at H.
The anticomplement of KH is the (parallel line) DL, tangent to D at L.

We have a special case of a general theorem for rectangular hyperbolas :

Theorem A
If a chord UV of a rectangular hyperbola is perpendicular to the tangent at W,
then UW and VW are perpendicular.

This in turn is a limiting case of another general theorem which is very easy to prove.

Theorem B
Suppose that UV and WZ are perpendicular chords of a rectangular hyperbola H.
Then U,V,V,Z is an orthocentric system (e.g. Z is the orthocentre of ΔUVW).

Since H is a rectangular hyperbola, and a circumconic of ΔUVW, it must pass through
the W', orthocentre of the ΔUVW. As UV and WZ are perpendicular, W' is on WZ. As
H meets WZ only at W and Z. Thus W = W', the orthocentre.

Proof of Theorem A
Consider the limiting case as Z approaches W along the curve. Then ZW tends to the
tangent at W. In the limit, ΔUVW has orthocentre W, so UW and VW are perpendicular.

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