1. Complements and anticomplements.
The complement of a point P is the mid-point of P and its anticomplement.
The centroid G = [1,1,1] lies on the line through these points.
Further, (P,P-,P+,G) = -1, so we have a harmonic range.
Given a point P = [p,q,r], the complement is P- = [q+r,r+p,p+q],
the anticomplement is P+ = [(q+r-p),(p+r-q),(p+q-r)].
If P is normalized, so that p+q+r=1, then so is P+.
Then the mid-point is just the "barycentric sum", so is P-.
The line PP- has equation (q-r)x+(r-p)y+(p-q)z = 0.
This clearly contains P+ and G.
Now P+ = -P + P-, G = P + P-, so we get the cross-ratio.
Note that we also have (G,P+,P-,P) = -1.
2. Conics with common perspector.
If an inconic I and a circumconic C have common perspector P,
then
P and the centres, I, C, of the conics are collinear.
The barycentric square P2 also lies on the line of centres.
Further (P,C,I,P2) = -1, so we have a harmonic range.
Say P = [p,q,r]. Then the centre of I is I = [p(q+r),q(r+p),r(p+q)].
The centre of C is C = [p(q+r-p),q(r+p-q),r(p+q-r)].
The line PI has equation x(q-r)/p+y(r-p)/q+z(p-q) = 0.
Then it is trivial to check that C and P2 = [p2,q2,r2] lie on PI.
Let s = p+q+r, then P2 = (sP-C)/2, I = (sP+C)/2. This gives the
stated value for the cross-ratio.
If we use the barycentric product, then we can rewrite C = PP+,
I = PP- = (P-1)-.
3. Harmonic ranges
Since complementation and anticomplementation are projective,
they preserve cross-ratio and hence harmonic ranges.
If we fix a point K, then mapping P to KP is also projective,
so preserves collinearity and cross-ratio. We can view (2) above
as a special case of (1), with multiplier P.
4. The centre of a circumellipse.
If C is a circumellipse, with centre C and perspector P, then
A point of the form P with U+V+W = 2Kπ lies inside the Steiner
Thus C is a circumellipse if and only if the centre C has the form
C = [sin(U),sin(V),sin(W)], P = [sin2(U/2),sin2(V/2),sin2(W/2)],
where U+V+W = 0.
Inellipse, so is the
perspector of a circumellipse.
[sin(U),sin(V),sin(W)] with U+V+W = Kπ.
Note. We may replace U, V, W by U+kπ, V+lπ, W+mπ with k,l,m
all odd or all even to get "angles" which sum to any multiple of 2π.
If the multipliers are odd, this reverses the sign of all of the
coordinates of C.
Examples are the circumcircle with U=2A,V=2B,W=2C, and the
Steiner
Ellipse with U=V=W=2π/3.
The proofs are non-trivial. you can find them here.
5. Common centre and asymptote.
Let L be the line xu+yv+zw = 0. Suppose that there are a
concentric inconic I and circumconic C with L as asymptote.
Then U = [u,v,w] lies on the Steiner Ellipse.
The other asymptote of C is L-1 : x/u+y/v+z/w = 0.
The common centre is [(v-w)/u,(w-u)/v,(u-v)/w].
The perspector of C is [u(v-w)2,v(w-u)2,w(u-v)2].
The perspector of I is [1/u2(v-w),1/v2(w-u),1/w2(u-v)].
The other asymptote of I is L2 : xu2+yv2+z2 = 0.
Although the result is lengthy, most of the proofs are very short.
We know that the circumconic with asymptote L has the stated
centre and perspector. Its other asymptote is L-1.ref
We also know that the inconic with asymptote L is concentric with
the circumconic having as asymptotes the lines xf+yg+zh = 0
and x/f+y/g+z/h = 0, with
[f,g,h] = [1/v+1/w-1/u,1/w+1/u-1/v,1/u+1/v-1/w].
One of these must be L. If it is the former, then we soon see that
u=v=w, so L is the line at infinity which cannot be an asymptote.
If it is the latter, f = k/u, g = k/v, h = k/w for some k. Then either
k = 1, and u = v = w, which we saw is impossible, or we must have
1/u+1/v+1/w = 0. Thus U is on S.
Note that, as 1/u+1/v+1/w = 0, [f,g,h] = [1/u,1/v,1/w]. Then the
second asymptote of I is L2. The perspector is as stated.ref
Note. G = [1,1,1] lies on L-1.
6. Circumconics with common tangent.
Let L be the line ux+vy+wz = 0.
The circumconics touching L have perspectors on the inconic
with perspector U-1 = [1/u,1/v,1/w].
The circumconic which touches L at P = [p,q,r] has perspector
Q = [up2,vq2,wr2]. ref
As P is on L up+vq+wr = 0. It is easy to check that Q lies on
the stated inconic.
Note. The inconic is the dual of the isotomic conjugate of L.
7. Circumparabolas
If the circumparabola C touches the line at infinity at U = [u,v,w],
then C has centre U2 = [u2,v2,w2] which lies on the
Steiner Inellipse
and perspector U.
These results are just a special case of earlier results for circumconics. ref
The remark about the location of U2 follows from 6 since here the line
is x+y+z = 0, so the inconic has perspector [1,1,1]. This is the Steiner
Inellipse.
8. A result related to 4.
Suppose that u+v+w = π. Then the inconic I with centre
Q = [sin(u),sin(v),sin(w)] has perspector P' = [tan(u/2),tan(v/2),tan(w/2)].
We know that P' = (Q+)-1. The result follows from the simple calculation
sin(v)+sin(w)-sin(u) = 4cos(u/2)sin(v/2)sin(w/2), and similar considerations
for the y- and z-coordinates.
In the notation of 4, we make the choice of U,V,W so that U+V+W = 2π.
Then put 2u = U, 2v = V, 2w = W, so that the above result can be applied.
Note that the centre Q is a square root of the perspector P in 4.
If the circumconic C of 4 is the circumcircle, then Q = [a,b,c]. Thus the
inconic I is the incircle, and the perspector is [tan(A/2),tan(B/2),tan(C/2)],
the Gergonne point.
If the circumconic C of 4 is the Steiner Ellipse, then Q = [1,1,1]. Thus the
inconic I is the Steiner Inellipse, and the perspector is again Q, the centroid.
9. Some more circumellipses.
We begin with an example :
The incentre I can be written as [sin(A+π),sin(B+π),sin(C+π)].
By 4, the circumconic C1 with centre I is an ellipse as the angle sum is 4π.
As in 4, the perspector can be written as the barycentric square J2,
where J = [sin((A+π)/2,sin((B+π)/2),sin((C+π)/2)]. Now J has angle sum
equal to 2π, so is the centre of a further circumellipse C2. By 4 this has
perspector K2, where K = [sin((A+π)/4,sin((B+π)/4),sin((C+π)/4)].
This has angle sum π, so we add π to each angle to allow us to proceed.
Observe that the adjusted version of K has angle sum 4π, the same as I.
Thus we may proceed to define a sequence of circumellipses. At every
other stage, we need to add π to each angle.
Notes.
The perspector of C1 is the Mittenpunkt, X(9) in ETC.
J = [cos(A/2),cos(B/2),cos(C/2)], so is the point X(188).
The perspector of C2 may be found from 4. It is X(236).
As we have seen this is the square of the centre of a further
circumellipse C3.
The incentre arises in this way from the circumcircle C0.
In turn, C0 arises from the circumellipse with centre
X(1147) = [sin(4A),sin(4B),sin(4C)].
We now have a chain of four circumellipses with known centres.
Of course, we see that we can continue backwards along the
chain with centres [sin(2nA),sin(2nB),sin(2nC)].
We have used nothing special about I except that it is the centre
of a circumellipse.
Suppose that C is a circumellipse with centre C and perspector P2.
As n tends to infinity the C(n) tend to the Steiner Ellipse.
Then we can define a sequence of circumellipses {C(n)} as follows:
C(1) is C,
C(n+1) is the circumellipse with centre C(n+1) = P(n), where
P(n)2 is the perspector of C(n).
The example shows the heart of the argument. We begin by writing C in the
form [sin(U),sin(V),sin(W)], where U+V+W = 2kπ. We can choose U,V,W in
the range (0,2π), so that k is 1 or 2. The detail is slightly different in these
cases, but the reader should be able to make the necessary adjustments
for the case k = 1.
Suppose that k = 2. As in the example, we create the sequence.
We will calculate the first angle U(n) only. The others are similar.
Observe that, if n is even, then the angle sum is 4π, and then
U(n+1) = U(n)/2. Now the angle sum is 2π, so U(n+2) = U(n+1)/2+π
and the angle sum is 4π again.
If we concentrate on the even values of n, we see that we have
U(2m) = U/22m + π(1 + 1/4 + ... + 1/4m-1).
As m tends to infinity, U(2m) tends to 4π/3.
Also U(2m+1) = U(2m)/2, so these tend to 2π/3.
Thus in either case, the centres tend to the centroid G = [1,1,1]
and the circumellipses to the Steiner Ellipse.
10. The centre and perspector of a circumconic.
The circumconic with perspector P = [p,q,r] has centre
C = [p(q+r-p),q(r+p-q),r(p+q-r)].
This is the X(2)-Ceva conjugate of P.
There are 21 examples of such pairs in ETC.
These include (X1),X(9)), (X(188),X(236)) we met in 9.
Other examples are (X(2),X(2)), (X(3),X(6)), (X(4),X(1249))
and (X(5),X(216)).
11. The intersection of circumconics.
If circumconics have perspectors [p,q,r] and [p',q',r'], then
the fourth intersection is [1/(qr'-q'r),1/(rp'-r'p),1/(p'q-pq')].
This is just the observation that the conics are the isotomics of
the lines px+qy+rz = 0 and p'x+q'y+r'z = 0. The intersection
of the conics other than X,Y,Z is the isotomic conjugate of the
intersection of the lines.
If we take the circumcircle and the circumellipse with centre X(1),
the perspectors are X(6) and X(9), so the intersection is X(100).
For the circumcircle and the circumellipse with centre X(1147),
the fourth intersection is X(110).
12. A sequence of intersection points.
Suppose we have the sequence {C(n)} with C(1)a given ellipse.
If n > 1, say C(n) has centre [sin(U),sin(V),sin(W)]. Then the centre
of C(n-1) can be written as [sin(2U),sin(2V),sin(2W)]. By 11,
the fourth intersection of these conics is the isotomic conjugate of
the point with x-coordinate sin(V)sin(2W) - sin(2V)sin(W). This can
be rewritten as 2sin(U)sin(V)sin(W)sin((V-W)/2)/sin(U/2). Now we
can remove the symmetrical factor to get sin((V-W)/2)/sin(U/2).
As n tends to infinity, U/2 approaches 2π/3 or π/3 which have the
same sine. Also, from 11, (V-W) = (V*-W*)/2n, where the centre
of C(1) has angles U*,V*,W*. If we consider the ratios of the
coordinates and use the result that sin(αx)/sin(βx) tends to α/β.
as x tends to zero, the fourth intersection tends to the point
[1/(V*-W*),1/(W*-U*),1/(U*-V*)].
When C(1) is the circumcircle, the limit point is [1/(B-C),1/(C-A),1/(A-B)].
13. Some centres of circumellipses.
We have seen that X(1), X(2), X(3), X(188) are the centres of circumellipses.
We can check that the following also qualify:
The Mittenpunkt X(9) = [a(b+c-a),b(c+a-b),c(a+b-c)], perspector X(1).
The Speiker centre X(10) =[b+c,c+a,a+b], perspector X(37), and
X(37) = [a(b+c),b(c+a),c(a+b)], perspector X(10).
In each case, we check that the corresponding perspector is inside the
Steiner Ellipse.
Cabri shows that X(4), X(5), X(6), X(7), X(8) do not always give ellipses.
14. An identity for the X(2)-Ceva conjugate and others.
A straight-forward calculation shows that the X(2)-Ceva conjugate can be
expressed as the complement of the isotomic conjugate of the anticomplement.
Thus, the X(2)-Ceva conjugate of U is ((U+)-1)-.
This makes it clear why the transform has order 2.
In a similar way, the P-Ceva conjugate of U is constructed as follows
P (((U/P)+)-1)-, again an algebraic conjugate of
an isotomic conjugate, and
hence of order 2.
The Cevapoint of P and U can be written P((U/P)-)-1 or
U((P/U)-)-1.
Either way it is a congugate of the map sending Q to (Q-)-1.
The crosspoint of P and U can be written P((U/P)-1)- or
U((P/U)-1)-.
Either way it is a congugate of the map sending Q to (Q-1)-.
15. The Steiner ellipses.
We know that the Steiner Inellipse S' is the complement of the Steiner Ellipse S.
Suppose that U =[u,v,w] lies on S. Then U- = [v+w,w+u,u+v] lies on S'.
Let G = [1,1,1]. Normalizing as necessary, UG/GU- = 2, Thus U- is the point
on S' "diametrically opposite" U. The other intersections with UG are given by
[2v+2w-u,2u+2w-v,2u+2v-w] on S and [4u+v+w,u+4v+w,u+v+4w] on S'.
As above [P,P-,G,P+] = -1, so PG/GP+ = 1/2.
The point P lies inside S' if and only if its anticomplement lies inside S.
16. Locating the centres of circumellipses.
In 4, we gave an algebraic characterization of the centre of a circumellipse.
We also gave a geometrical description of the perspector as an interior point
of the Steiner Inellipse S'. Here we find the regions in which the centre lies.
We know that U = [u,v,w] is the centre of a circumellipse if and only if
the perspector U* = [u*,v*,w*] lies within S', i.e. F(u*,v*,w*) < 0, where
F is given in 4. We have u* = u(v+w-u), v* = v(w+u-v), w* = w(u+v-w),
so that F(u,v,w) = -(u*+v*+w*).
Now we see that the condition that U be the centre of a circumellipse can be
expressed as u**+v**+w** > 0.
Multiplying out, u**+v**+w** = (u+v+w)(u+v-w)(u+w-v)(v+w-u).
Note that F(U*) = F(U2).
As U is the centre of a circumellipse, U* lies in S' so can be taken
with positive coordinates. Then U* = V2. There are essentially four
choices of V. Then F(V*) = f(V2) = F(U*). Thus V is the centre of a
further circumellipse. Even if U is a centre, it is not clear that we can
choose V as a triangle centre. This relates to 4.
If we normalise U so that u+v+w > 0, then the condition can be expressed
as either (u+v-w), (u+w-v), (v+w-u) are all positive, or exactly two are
negative.
Consider the lines x+y-z = 0, x-y+z = 0, -x+y+z = 0. These are the lines
through the mid-points of two of the sides of ΔXYZ so parallel to the sidelines.
They divide the plane into seven regions.
If U is in the finite (triangular) region, then all of our factors are positive.
If U is in a region which contains a vertex, then exactly one factor is negative.
For example, in the region containing X, only (v+w-u) is negative.
Thus U will be the centre of a circumellipse if and only if it lies in the
interior of a region which does not contain a vertex of the triangle.