Although we use mainly projective methods, we want to apply our theory to
plane geometry, with concepts such
as ratios along a line, centre and
asymptote of a conic. To achieve this, we take the embedding of the
projective
plane on the plane E : x+y+z = 1. Points on L∞ :
x+y+z = 0 are points at infinity and L∞ the line at
infinity. Such
points define directions on E, i.e. a class
of parallel lines.
A point U =[u,v,w] not on L∞ embeds as the point U'
= (u/σ,v/σ,w/σ), where σ = u+v+w.
Provided we use only affine concepts, the
coordinates of U' can be viewed as normalized barycentric coordinates
for U. More generally, the coordinates of U are its barycentric
coordinates.
Of course, the barycentric coordinates can be replaced by
λu,λv,λw for any non-zero λ.
Definition
The centre of a plane conic is the pole
of the line at infinity.
Theorem 3
(1) The circumconic C with
C-perspector P = [p,q,r] has centre [p(q+r-p),q(p+r-q),r(p+q-r)].
(2)
The inconic I with I-perspector P = [p,q,r] has centre
[p(q+r),q(p+r),r(p+q)].
These follow at once from the matrices of the conics. A tedious calculation shows more :
Corollary 3.1
If the circumconic with perspector P has
centre Q, then the circumconic with perspector Q has centre P.
Some of the expressions which appear in Theorem 3 occur quite frequently, so we make the
Definitions
Given points U = [u,v,w], and U' =
[u',v',w']
the complement (or inferior) of U is the point
U- = [v+w,w+u,u+v],
the anticomplement (or superior) of U
is the point U+ = [v+w-u,w+u-v,u+v-w],
the isotomic
conjugate of U other than X,Y,Z is the point U-1 =
[vw,wu,uv].
the barycentric product of U and U' is the point U&U'
= [uu',vv',ww'].
Note that the complement and anticomplement are mutually inverse
projective transformations.
The complement maps X to [0,1,1], the
mid-point of YZ, and similarly for Y,Z, so that it maps the
reference
triangle to the triangle known as the medial triangle.
If uvw is non-zero, we
may write the isotomic congugate of U as [1/u,1/v,1/w], from which it is
clear that isotomic conjugation is an involution (but not projective). As
the notation implies, the
isotomic conjugate is the inverse in the group
defined by the barycentric product. The identity is
the point [1,1,1], the
centroid of the reference triangle.
We can restate our results for conics as follows :
Theorem 3'
If the inconic I has
I-perspector P, then its centre is (P-1)-.
If
the inconic J has centre Q, then it has J-perspector
(Q+)-1.
If the circumconic C has
C-perspector P, then its centre is P&P+.
If the
circumconic D has centre Q, then it has D-perspector
Q&Q+.
If we use the section formula in the embedding, it is easy to solve the following
Exercise
For any point not on the line at infinity,
P+ is the mid-point of PP-.
We know that, given four points and a line through one of the points, there
is in general a unique conic through the
points with the line as tangent. We
want a quantified form.
Theorem 4 Proof
Suppose that L : ux+vy+wz = 0 does
not pass through a vertex of the reference triangle T*,
and P =
[p,q,r] is a point on L, but not on any sideline of T*.
Then
(1) the unique circumconic C touching L at P has
C-perspector [up2,vq2,wr2]
(2) the
unique inconic I touching L at P has I-perspector
[1/pu2,1/qv2,1/rw2],
The centre of I
is
[qv2+rw2,pu2+rw2,pu2+qv2]
If the circumconic C has C-perspector is Q = [α,β,γ],
and P is on C, the tangent at P is αx/p2 + βx/q2 +
γx/r2 = 0.
This is L if and only if Q =
[up2,vq2,wr2]. We can check that, for this Q, P
is on C as P lies on L.
For (2), we consider the dual problem.
This requires a circumconic C' touching L' : px+qy+rz = 0
at
the point P' = [u,v,w] (which lies on L' as P is on L). By (1),
this has C'-perspector
[pu2,qv2,rw2]
Since I is the dual of
C', we have the result about the I-perspector.
The centre of
I follows from Theorem 3.
We will be particularly interested in the case where L is an asymptote
to the conic. Then P is the
intersection of L and L∞. Thus, P =
[v-w,w-u,u-v]. It turns out that the centres of the conics are
quite simple
in this case.
Theorem 4'
Suppose that L : ux+vy+wz = 0 does
not pass through a vertex of the reference triangle T*,
and is not
parallel to any sideline of T*. Then
(1) the unique circumconic
C with asymptote L has C-perspector
[u(v-w)2,v(w-u)2,w(u-v)2],
centre
[u(v2-w2),v(w2-u2),w(u2-v2)].
(2)
the unique inconic I with asymptote L has I-perspector
[1/(v-w)u2,1/(w-u)v2,1/(u-v)w2],
centre
[(uv+uw-vw)(v-w),(vu+vw-uw)(w-u),(wu+wu-uv)(u-v)].
The point F = [uv+uw-vw,vu+vw-uw,wu+wu-uv] whose coordinates arise above will
appear later.
For the moment, we note that F = (U-1)+,
where U = [u,v,w].
Note.
We observe that the point U2 =
[(v-w)/u2,(w-u)/v2,(u-v)/w2] also lies on
I.
By considering the dual, the tangent to I at U2 is
u2x+v2y+w2z = 0.
A theorem of newton
If a line L cuts the
sidelines of triangle T = ΔABC at points P on BC, Q on CA, R on AB,
then the mid-points of AP,BQ and CQ are collinear in a line TL. If
I is an inconic of T
which touches L, then the centre of
I lies on TL.
The line TL is the newton line for T and L. We prove this by coordinates, taking the reference triangle as T.
Theorem 5 Proof
Suppose that L : ux+vy+wz = 0 does
not pass through a vertex of the reference triangle T*,
and is not
parallel to any sideline of T*. Let the line L meet x=0 at P, y=0
at Q, z=0 at R.
(1) The mid-points of XP,YQ,ZR lie on the line TL:
(uv+uw-vw)x+(vu+vw-uw)y+(wu+wu-uv)z = 0.
(2) Any inconic I of
T* touching L has its centre on TL.
The line L meets x=0 in P = [0,w,-v]. The normalised form is
therefore [0,w/(w-v),-v/(w-v)].
Thus the mid-point of XP is [w-v,w,-v].
Similarly the other mid-points are [-w,u-w,u], [v,-u,v-u].
Clearly, these are
the rows of a singular matrix, so the points are collinear. Using any two,
we
get the equation of TL as stated.
Suppose that I touches
L at [p,q,r]. The centre of I is given in Theorem 4. If we put
these
in the equation for TL, we get uvw(pu+qv+rw) = 0, as [p,q,r] is
on L.
inparabolas
The family of inconics of T touching L contains one
parabola since a parabola touches L∞.
A parabola is a conic with
centre on L∞, so our result on the centre of I quickly
yield:
Theorem 6
Proof
Suppose that L : ux+vy+wz = 0 does
not pass through a vertex of the reference triangle T*,
and is not
parallel to any sideline of T*. The inparabola I of T*
touching L has contact point
[vw(v-w),uw(w-u),uv(u-v)]. It has centre
Q = [u(v-w),v(w-u),w(u-v)] and I-perspector Q-1.
From Theorem 4, the centre is on L∞ if and only if
pu2+qv2+rw2 = 0,
where P = [p,q,r] is the
contact point with L. As P is on L, pu+qv+rw = 0.
These two
equations determine P as [vw(v-w),uw(w-u),uv(u-v)].
Again by Theorem 4, the
inconic has the required I-perspector and centre.
a coincidence?
Theorem 7 Proof
Suppose that L : ux+vy+wz = 0 does
not pass through a vertex of the reference triangle T*,
and is not
parallel to any sideline of T*. Let TL be the line defined in
Theorem 5.
Then the circumconic with asymptote TL is concentric with
the inconic with asymptote L.
>From Theorem 4', the centre of the inconic is
[(uv+uw-vw)(v-w),(vu+vw-uw)(w-u),(wu+wu-uv)(u-v)].
For the circumconic, we
again apply Theorem 4', but we must replace u,v,w by the coordinates of
F.
The calculation is rather tedious, but we find the same centre. The crux
is that, if F = [f,g,h], then
f(g+h)(g-h) = 2fvw(uv-uw) = 2uvwf(v-w).
Note that, if we have an inconic which is a hyperbola, then an asymptote
defines a line L. From this
we find the line TL and the
concentric circumconic is also a hyperbola. Equally, we can start with
a
circumconic which is a hyperbola. We can take either asymptote as TL. We
have observed that the
coefficients of the equation give the point F. We can
then recover the line L as that whose coefficients
come from
(F-)-1. Thus we recover the inconic which must be a
hyperbola.
Note also that the common centre is the intersection of L and TL.
We can easily compute the C-perspector of the circumconic concentric
with the inconic. Using
the results of Theorem 4' as above we find that is
[u2(v-w)2f,v2(w-u)2g,w2(u-v)2h],
where F = [f,g,h].
In abbreviated notation, it becomes
U&U&U'&U'&(U-1)+, where U' =
[v-w,w-u,u-v].
We can now identify the second asymptote of C. It is easy to check
that U&U'=[u(v-w),v(w-u),w(u-v)]
lies on the line TL : fx+gy+hz =
0. The point is also on L∞ , so must be the contact point with C.
But then U&U'&F lies on C, from the equation for C,
and on L∞, so gives the second asymptote as the
tangent here. Using
the formula for a tangent, the asymptote is x/f+y/g+z/h = 0. This is the dual
of
F-1. As indicated above, we can recover the corresponding
asymptote for I by finding the isotomic
conjugate of the complement of
this point. This calculation gives the second asymptote as the line
L'
: ux/f+vy/g+wz/h = 0, with contact point (where this meets L∞)
[(v-w)f2,(w-u)g2,(u-v)h2] = U'&F&F.
Example Solution
If C and I are, respectively,
the circumconic with C-perpsector P and the inconic with
I-perspector Q
and are concentric at R, then P&Q=R.
>From Theorem 3', we have Q = (R+)-1,
and P = R&R+. The result follows.
a related circumconic
from Corollary 3.1, we know that the circumconic C with
C-perspector I is related to the circumconic
C* with centre P.
For the circumconic C of Theorem 7, C* has C*-perspector
[f(v-w),g(w-u),h(u-v)],
so passes through the points F, U-1,
F&U-1 and F-1. The calculation for the last is
tedious. The tangents
are the duals of U'&F-1,
U&U&U'&F-1, U&U&U-1&F-1
and F&F&F&U'.
note on poristic families
Recall that, if triangle T has circumconic C and inconic
I, then the poristic family F(T,C,I) consists of the
set of triangles with circumconic C and inconic I. Provided we
work over complex numbers, there is one
from each point of C. For our
case of concentric hyperbolas, we see that the family contains the
reflection
of T* in the common centre. This is homothetic with T*, factor -1.
Note that, if L is an asymptote
and is geometrically related to
T*, then it will have the same geometrical relationship with the
reflection.
For example, if L it the euler line of T*,
that is the line through the centroid and circumcentre, then will be
the
euler line of the reflection. This is the classic example, the reflected
triangle is the Gossard Triangle.
The centre of the homothety if the
Gossard Perspector. We have identified it as the common centre of two
conics,
and as the intersection of the euler line and its associated newton line.We use
the notation of ETC.
L is the euler line.
U = X(525),
U'= X(30),
F =
anticomplement of X(648),
The centre is X(402).
Further examples
There are two cases where we can identify the common centre associated with
simple central lines.
(1) L : x/a+y/b+z/c = 0, the antiorthic axis (2) L : x/a2+y/b2+z/c2 = 0, the
lemoine axis
U = [1/a,1/b,1/c] =
X(75),
U'= [a(b-c),b(c-a),c(a-b)] = X(513),
F = [b+c-a,c+a-b,a+b-c] =
X(8), the Nagel point,
TL is the gergonne line, infinite point
X(514).
I-perspector = X(100).
C-perspector = X(11), the
feuerbach point, the centre of the feuerbach hyperbola.
The centre is
U'&F = X(650) (so this is the perspector of the feuerbach
hyperbola).
The second point on C is U&U'&F = X(522). This
identifies the second asymptote as
the line joining X(650) and X(522). It is
the dual of F-1 = X(7).
The second point on I,
U'&F&F is not related to any point of ETC.
The second asymptote of
I is the dual of U&F-1 = X(85) = X(9)-1.
Notes.
F = X(8), U-1 = X(1), F-1 = X(7) and
F&U-1 = X(9) lie on the feuerbach hyperbola.
The yff parabola
touches L at U-1&U' = X(649). It has centre X(514) on
TL.
U = [1/a2,1/b2,1/c2] =
X(76),
U'=
[a2(b2-c2),b2(c2-a2),
c2(a2-b2)] = X(512),
F =
[b2+c2-a2,c2+a2-b2,
a2+b2-c2] = X(69).
TL is the orthic
axis, infinite point X(523).
I-perspector =
X(110).
C-perspector = X(125), the centre of the jerabek
hyperbola.
The centre is U'&F = X(647) (so this is the perspector
of the jerabek hyperbola).
The second point on C is U&U'&F =
X(525).This identifies the second asymptote as
the line joining X(647) and
X(525). This includes X(441) and X(905). It is the dual
of F-1 =
X(4).
The second point on I, U'&F&F is the isotomic conjugate
of X(107).
The second asymptote of I is the dual of
U&F-1 = X(264) = X(3)-1.
Notes.
F = X(69),
U-1 = X(6), F-1 = X(4) and F&U-1 = X(3) lie
on the jerabek hyperbola.
The kiepert parabola touches L at
U-1&U' = X(669). It has centre X(523) on TL.
These examples are related. From an earlier note, the line L of (2) is
a tangent to I of (1).
It follows that the centre of the inconic of
(1) lies on the line TL of (2), i.e. X(650) is on the
orthic axis.
Thus, the orthic, antiorthic and gergonne axes concur at this point.