Date: Jun 30, 1998 9:22 PM
Author: steve sigur
Subject: Conway on Trilinear vs Barycentric coordinates

 In response to frequent mention of Trilinear coordinates on this 
newsgroup, John Conway admonished us that Barycentric coordinates were to
be preferred. He sent some of us the following comparison of the two.
Since my next post will comment on this, I thought I would post it first
in its unadulterated form. I have learned much Geometry studying this.

Steve Sigur


On 12/18/97 john Conway wrote

I'm sending this to several people interested in triangles, and
hope they'll confirm receipt (except for rkg, who might be out of
email touch), and send any further substantial messages about triangles
to everyone else on the list.

There are three standard systems that use three coordinates to
represent a point in the plane of a give triangle, namely


Often the last is shortened to "TRILINEARS", but I prefer the longer
name since in fact all three systems are trilinear. Each may or may
not be normalized. For the not-necessarily normalized versions I'll use:

(X:Y:Z) for B and A, and [x:y:z] for OT

and for the normalized ones

(X,Y,Z) for B, and [x,y,z] for OT.

All three (unnormalized) systems are very similar - indeed they
for Barycentrics and Areals, so I'll usually call these jointly BA,
and the conversion from TO to these is very simple:

[x:y:z] becomes (ax:by:cz).

So for many purposes it hardly matters which system one uses. However,
there ARE ways in which one or other of the systems is better than
and it is the purpose of this note to point out that when one takes all
these into account the barycentric system emerges as the clear winner.

The letters B,A,BA,OT before each numbered point below show how this
decision was reached. A letter N indicates that normalized coordinates
are involved.

OT 0N. The distances of P from the sides are

2X.Delta/a, 2Y.Delta/b, 2Z.Delta/c [ x, y, z ].

A 1N. The areas of PBC, PCA, PAB are the normalised areal coordinates

X.Delta, Y.Delta, Z.Delta [ ax/2, by/2, cz/2 ]

B 2N. If VA, VB, VC are vectors to A, B, C , then

P = X.VA + Y.VB + Z.VC

[Of course these are just the definitions of the three systems.]

BA 3. The Cevian ratios are

Y:Z, Z:X, X:Y [ by:cz, cz:ax, ax:by ].

BA 4. The Menelean ratios of the lines

PX + QY + RZ = 0 and px + qy + rz = 0
-Q:R -R:P -P:Q and -cq:br -ar:cp -bp:aq

B 5. The normalizing condition is

X + Y + Z = 1 (B) or Delta (A), [ ax + by + cz = 2Delta ]

OT 6. The isogonal conjugate (or "conjugal") is

co-P = ( a^2/X : b^2/Y : c^2/Z ) [ 1/X : 1/Y : 1/Z ]

BA 7. The isotomic conjugate (or "isotome") is

iso-P = ( 1/X : 1/Y : 1/Z ) [ 1/(x.a^2) : 1/(y.b^2) : 1/(z.c^2)

B 8. AFFINE INVARIANCE. If an affine transformation takes

A, B, C, and P = (X,Y,Z) to A1, B1, C1 and P1,

then the barycentric coordinates of P1 with respect to
the new triangle A1 B1 C1 are still (X,Y,Z).

(It is because isotomic conjugation is an affinely invariant concept
that its expression (see #7) in barycentrics cannot involve the edge
lengths of the triangle.)

BA 9. The concepts of subordinate and superior points
sub-P and super-P are particularly important in the theory.
These points are the images of P in the subordinate
(or "medial") triangle, whose vertices are the midpoints
of the edges of ABC, and the superior (or "anticomplementary")
triangle, the midpoints of whose edges are A,B,C. We have:

sub-P = ( Y+Z : Z+X : X+Y ) [ (by+cz)/a : (cz+ax)/b : (ax+by)/c ]
super-P = (Y+Z-X:Z+X-Y:X+Y-Z)

(Again the simplicity of the BA coordinates is due to affine invariance.)

B 10. RATIONALITY. X,Y,Z are rational functions of the
Euclidean coordinates of the points A,B,C,P. If P is
a point that's rationally defined from A,B,C, then its
barycentric coordinates are rational functions of
a^2, b^2, c^2. This is true, for instance, of the
centroid, orthocenter, circumcenter, symmedian point,
Brocard points, and so on.

(This is an extremely important point, and extends to give the very
useful property below.)

BA 11. ALGEBRAIC CONJUGATES. Many points (such as the incenter)
can be obtained by solving algebraic equations that have
other (algebraically conjugate) solutions. Passing to these
other solutions then yields further points that have
essentially the same geometric properties (in this way, we
get from the incenter to the excenters). We can get the
barycentric coordinates of such "companions" as the appropriate
algebraic conjugates of those of the original.

The simplest case of this is when the coordinates are
rational functions of a,b,c but not of a^2, b^2, c^2.
So for example if a point that's rationally constructed from the
incenter has barycentric coordinates

( X(a,b,c), Y(a,b,c), Z(a,b,c) )

then the corresponding point obtained from the a-excenter is
simply obtained by "changing the sign of a", thus:

( X(-a,b,c), Y(a,-b,c), Z(a,b,-c) ).

For example the Nagel point is the

"super-incenter" (b+c-a : c+a-b : a+b-c),

and so its a-companion is (b+c+a : c-a-b : -a+b-c).

The OT coordinates of these points are much harder to understand:

[ b/a + c/a - 1 : c/b + a/b - 1 : a/c + b/c - 1]
[ b/a + c/a + 1 : c/b - a/b - 1 : -a/c + b/c - 1].

Well, that will do for now. I'll just survey the "winners"

0 1 2 3 4 5 6 7 8 9 10 11

In only two cases is OT the winner, and in all other cases
but one (the definition of A!) B is at least a joint winner.

I have deliberately preferred conceptual reasons for preferring
one system to another, rather than mere comparisons of the simplicity
of the coordinates for particular points. Some points are simpler
under one system rather than another, and often it's OT that would give
the simpler ones. But this difference can never be great, since
[x,y,z] translates to [ax,by,cz]; and in barycentrics the simplicity
often has a useful conceptual meaning.

For example (1:1:1) = [ 1/a : 1/b : 1/c ] is the centroid,
and its simplicity in barycentrics comes from its affine invariance.

On the other hand (a:b:c) = [1:1:1] is the incenter, more
complicated in barycentrics since it has algebraic conjugates
(-a:b:c) (a:-b:c) (a:b:-c).
The apparent simplicity in trilinears disappears when we pass to the
sub-incenter (Spieker point)

(b+c:c+a:a+b) = ( (b+c)/a : (c+a)/b : (a+b)/c ).

The "Morley perspectors" look simpler in OT:

[ cos(A/3) : cos(B/3) : cos(C/3) ] and [ sec(A/3) : sec(B/3) : sec(C/3) ]

but again this apparent simplicity disappears when we want a bit more:
the Barycentric versions

( a.cos(A/3) : ... ) and ( a.sec(A/3) : ... )

can be conjugated to get all the "companion Morley perspectors"
just as easily.

In summary, the simplicity of coordinates for particular points can go
either way, and in any case is not a strong argument. It's their
properties that make barycentrics the clear winner.

John Conway