Conway on Trilinear vs Barycentric coordinates
steve sigur
I have learned much from John Conway's discussion of coordinate systems
natural to triangles, but some reasons that I like Trilinear Coordinates
were not mentioned. Although I am trying to use barycentric coordinates
as much as possible, I still find myself going back to Trilinears for
reasons that are not on Conway's list.
First
Duality (this is related to the algebraic conjugacy discussed below):
With Trilinear coordinates for points come trilinear equations for lines.
I know there are barycentric lines as well, but trilinear lines can be
written in a form that gives them a dual role computing distances to
points. I find this very powerful. In fact I prefer to do geometry
positing lines first, rather than points first. It seems to me that
trilinear lines are better.
Second
I think that that Isogonal conjugacy (reflection over bisectors), which
involves many of the major centers in triangles, is more significant that
isotomic conjucacy (over medians). Isogonal conjugacy is nice for
trilinear points and lines.
John Conway wrote
> 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.
>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]
> and
> [ b/a + c/a + 1 : c/b - a/b - 1 : -a/c + b/c - 1].
This is exactly one of the arguments I would make in favor of trilinears.
This sort of thing is easy and natural in trilinears. If P is a point
inside the triangle whose sides have equations in trilinear form alpha =
0, beta = 0, gamma = 0, then the cevians have equations
m alpha - n beta = 0, n beta - p gamma = 0, p gamma - m alpha = 0
where the coefficients m , n and p are functions of the sides or angles
of the triangle.
These lines are concurrent at P whose trilinear coordinates are (1/m,
1/n, 1/p). [Note that (alpha, beta, gamma) = (1/m, 1/n, 1/p) solves the
above three equations. It is this ability of a line alpha = 0 to also
become distance alpha, that gives a computational boost to trilinear
coordinates and lines].
The equations for the external cevians (also called harmonic conjugates
lines) are
m alpha + n beta = 0, n beta + p gamma = 0, p gamma + m alpha = 0.
These, taken in pairs along with the appropriate internal cevian, are
concurrent at the points
(-1/m, 1/n, 1/p), (1/m, -1/n, 1/p), (1/m, 1/n, -1/p).
These, I believe, are the points corresponding to (1/m, 1/n, 1/p) in the
same way that excenters are algebraically conjugate to the incenter.
Steve Sigur
steve sigur
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
BARYCENTRICS (B), AREALS (A), and ORTHOGONAL TRILINEARS (OT).
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
coincide
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
another,
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
[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
are
-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)
[(by+cz-ax)/a:(cz+ax-by)/b:(ax+by-cz)/c]
(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]
and
[ 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
OT A B BA BA B OT BA B BA B BA
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
theoretical
properties that make barycentrics the clear winner.
John Conway
Kirby Urner
I read all of this and still can't fathom which of
these coordinate system types is this:
(0,0,1)
|
| * P
|
/ \
/ \
/ \
(1,0,0) (0,1,0)
Any point P is reachable by linear combination of
basis vectors. Only positive scalars required
(vector reversal operation not required to span
plane, only grow/shrink and tip-to-tail addition).
The 4 basis vector analog of the above is what
I've dubbed "quadrays" and describe at my website
at http://www.teleport.com/~pdx4d/quadrays.html
-- 4 vectors from the origin thru the vertices
of a regular tetrahedral enclosure. All points
in volume uniquely identified by 4-tuples with
no negative scalars.[1]
Kirby
[1] we can define negative scalars and include
vector reversal i.e. multiplication of a vector
by -1, but these more provide redundant shortcuts
in notation than trully novel operations.
steve sigur
>
>I read all of this and still can't fathom which of
>these coordinate system types is this:
>
> (0,0,1)
> |
> | * P
> |
> / \
> / \
> / \
> (1,0,0) (0,1,0)
>
>
>Any point P is reachable by linear combination of
>basis vectors. Only positive scalars required
>(vector reversal operation not required to span
>plane, only grow/shrink and tip-to-tail addition).
>
I do not think your system is either. Neither trilinears nor barycentric
coordinates represent distances from a common origin. Yours looks like
the regular basis vectors in 3D space. If yours is in 2D space then I
think it is different from the other two.
Whenever you describe a two dimensional idea (a point) interms of 3
numbers, you are giving up something important, namely that your basis
vectors are orthogonal with an operation (the dot product) representing
that orthogonality. With barycentrics and trilinears (both homogeneous)
we gain something by doing this. Many of the properties of a triangle do
not depend on scale, for which homogeneous systems are nice. Also the
dual nature of points and lines comes out nicely with this. What
advantages does your system have?
steve
Kirby Urner
> I do not think your system is either. Neither trilinears nor
> barycentric coordinates represent distances from a common
> origin. Yours looks like the regular basis vectors in 3D space.
> If yours is in 2D space then I think it is different from the
> other two.
Yes, the example I gave is for mapping a surface. Three
vectors fan out from the origin with coordinates (1,0,0)
(0,1,0)(0,0,1). In volume, I would use 4 vectors and
4-tuples {1,0,0,0} where {} means "any permutation of
the enclosed numbers".
> Whenever you describe a two dimensional idea (a point) in terms
> of 3 numbers, you are giving up something important, namely
> that your basis vectors are orthogonal with an operation (the
> dot product) representing that orthogonality. With barycentrics
> and trilinears (both homogeneous) we gain something by doing
> this. Many of the properties of a triangle do not depend on
> scale, for which homogeneous systems are nice. Also the dual
> nature of points and lines comes out nicely with this. What
> advantages does your system have?
Quadrays have some useful properties. All centers of close
packed unit radius spheres in the fcc are linear combinations
of vectors {2,1,1,0} i.e. have positive integer coordinates.
Other primitive polys have integer coordinates also. The
'home base' tetrahedron defined by the 4 basis vectors has
edges of length 2 and a volume of 1 (by definition). The
unit volume tetrahedron divides evenly into the octahedron,
cube (face diag = 2) and rhombic dodecahedron (long face
diag = 2). Any tetrahedron defined by 4 fcc centers has a
whole number volume.
Mostly I use the quadrays apparatus as a philosophers' toy to
investigate concepts of "dimensionality" (I proffer this system
as "4D" even though it maps conceptual volume) and "linear
independence" (not necessarily the same as orthogonality).
Whereas in XYZ lingo we would say one of the basis vectors
is a linear combination of 3 others, this depends on a
change in orientation (vector reversal) being lumped in
with grow/shrink scaling, as another instance of "scalar
multiplication". In the quadrays system (and the flat
surface analog), no negative scalars are needed to span
vector space, although the "-" operator is still defined
i.e. -(1,0,0,0) is a vector pointing 180 degrees from
(1,0,0,0).
In XYZ (left handed), the +++ octant is not symmetrical
about the origin and signage is permuted to define 7
other quadrants. The apparent greater simplicity of
3-tuples is in trade-off with the all-positive economy
of this alternative 4-tuples game.
I say "game" and "toy" because I'm not trying to give
the misleading impression that I think quadrays are
anything but -- useful in their own context, but not
a replacement or substitute for anything that's already
well-established. The more games the better, is what
math is all about, no?
Kirby
References:
http://www.teleport.com/~pdx4d/quadrays.html
http://www.teleport.com/~pdx4d/quadphil.html
John Conway
On Tue, 30 Jun 1998, steve sigur wrote:
> I have learned much from John Conway's discussion of coordinate systems
> natural to triangles, but some reasons that I like Trilinear Coordinates
> were not mentioned. Although I am trying to use barycentric coordinates
> as much as possible, I still find myself going back to Trilinears for
> reasons that are not on Conway's list.
I'm very puzzled by this comment, because the message was all about
trilinear coordinates. It's obvious in many ways that Steve didn't
learn enough!
> First
> Duality (this is related to the algebraic conjugacy discussed below):
> With Trilinear coordinates for points come trilinear equations for lines.
> I know there are barycentric lines as well, but trilinear lines can be
> written in a form that gives them a dual role computing distances to
> points. I find this very powerful. In fact I prefer to do geometry
> positing lines first, rather than points first. It seems to me that
> trilinear lines are better.
All of this applies equally well to all systems of trilinear coordinates,
not just the orthogonal trilinears that Steve likes. So it gives no reason
for choosing either of the two major systems over the other.
> Second
>
> I think that that Isogonal conjugacy (reflection over bisectors), which
> involves many of the major centers in triangles, is more significant that
> isotomic conjucacy (over medians).
I'd express this my saying that isogonal conjugacy is more interesting
than isotomic. But it would be unwise to choose Orthogonal Trilinears
to make the formula for isogonal conjugacy a little bit simpler, when it
makes a hundred other formulae a lot more complicated.
> Isogonal conjugacy is nice for trilinear points and lines.
Look! BOTH systems treat BOTH isogonal and isotomic conjugacy pretty
nicely:
isogonal: ( a^2/X : b^2/Y : c^2/Z ) [ 1/x : 1/y : 1/z ]
isotomic: ( 1/X : 1/Y : 1/Z ) [1/x.a^2 : 1/y.b^2 : 1/z.c^2 ].
Now I agree that you do get simpler formulae for isogonal conjugacy
in the OT system - it was one of the 3 out of 10 points in which that
system beat BA. But the improvement is slight, and totally outweighed
by the many things that get MUCH worse.
> John Conway wrote
>
> > 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.
[Then I went on to mistype a formula - I'll correct it now : the
a-conjugate of ( X : Y : Z ) is
( X(-a,b,c) : Y(-a,b,c) : Z(-a,b,c).
So for instance if (X:Y:Z) is some point related to the incenter,
this will be the point correspondingly related to the a-excenter.]
> >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]
> > and
> > [ b/a + c/a + 1 : c/b - a/b - 1 : -a/c + b/c - 1].
The following few paragraphs are totally vitiated by the fact that
Steve hasn't understood that all the virtues he assigns to orthogonal
trilinear coordinates in them hold equally for all systems of trilinear
coordinates, including the barycentric ones! So they are not arguments
that favor one system rather than another.
> This is exactly one of the arguments I would make in favor of trilinears.
> This sort of thing is easy and natural in trilinears. If P is a point
> inside the triangle whose sides have equations in trilinear form alpha =
> 0, beta = 0, gamma = 0, then the cevians have equations
>
> m alpha - n beta = 0, n beta - p gamma = 0, p gamma - m alpha = 0
>
> where the coefficients m , n and p are functions of the sides or angles
> of the triangle.
>
> These lines are concurrent at P whose trilinear coordinates are (1/m,
> 1/n, 1/p). [Note that (alpha, beta, gamma) = (1/m, 1/n, 1/p) solves the
> above three equations. It is this ability of a line alpha = 0 to also
> become distance alpha, that gives a computational boost to trilinear
> coordinates and lines].
>
> The equations for the external cevians (also called harmonic conjugates
> lines) are
>
> m alpha + n beta = 0, n beta + p gamma = 0, p gamma + m alpha = 0.
>
> These, taken in pairs along with the appropriate internal cevian, are
> concurrent at the points
>
> (-1/m, 1/n, 1/p), (1/m, -1/n, 1/p), (1/m, 1/n, -1/p).
>
> These, I believe, are the points corresponding to (1/m, 1/n, 1/p) in the
> same way that excenters are algebraically conjugate to the incenter.
(*: see below - JHC)
As I said, these very valuable properties hold of ALL systems of
trilinear coordinates, so you won't lose them by changing to barycentrics.
My long letter ignored them for that reason - it concentrated ONLY on the
properties that distinguish the two main rival systems of trilinear
coordinates, and showed that almost all of them favor BA over OT,
very strongly.
Steve described me as "admonishing" people to use barycentric
trilinears rather than orthogonal trilinears. That makes me
sound like a disciplinarian, which I'm not. But of course, the
word really means "warning about", and I'll accept it in that
sense; I AM warning you that if you use orthogonal trilinears
you'll be missing out on lots of things that are obvious in
barycentric trilinears.
I starred the last line of Steve's letter, because it illustrates
this and also involves another misunderstanding. Steve points out
have certain points geometrically related to the point whose orthogonal
trilinear coordinates are [x:y:z] (his 1/m, 1/n, 1/p) have coordinates
[-x:y:z], [x:-y:z], [x:y:-z]. This is just as true in barycentric
trilinears; if the first point is (X:Y:Z) the other three are
(-X:Y:Z), (X:-Y:Z), (X:Y:-Z).
But he then goes on to say that these are related to [x:y:z] in
the same way that the excenters are algebraically conjugate to the
incenter. But that's not what I was talking about. In barycentric
coordinates, the incenter and excenters are
(a:b:c) (-a:b:c) (a:-b:c) (a:b:-c)
and they are indeed related by changing the signs of the coordinates.
Since this is true in each system, it gives no reason to prefer one
to the other.
The algebraic conjugacy I was talking about is much more valuable.
These four points are related by changing the signs of a,b,c, as
well! The Nagel point is the point of concurrence of the lines from
the vertices to the contact points of the incircle, and from its
barycentric coordinates are (-a+b+c:a-b+c:a+b-c) we immediately
deduce that the lines form the vertices to the contact-points of the
a-excenter meet at (a+b+c:-a-b+c:-a+b-c) - just change the sign of a!
This exemplifies my whole point. Steve understands the relationship
between [x:y:z] and [-x:y:z] (etc), and points out that it's easy to
see in orthogonal trilinears, and that that's one reason he prefers
to use those coordinates to barycentrics. He obviously didn't understand
that it was just as easy to see in barycentric trilinears, but that doesn't
matter very much; after all, he DID see it as a valuable property of a
coordinate-system. What's more important is that he DIDN'T understand
the relationship between (X:Y:Z) and ( X(-a,b,c):Y(-a,b,c):Z(-a:b:c) )
which is just as useful, and which he WON'T ever see unless he switches
to barycentric trilinears.
In summary: barycentric trilinears have all but one of the properties
that Steve adduces as reasons for prefering orthogonal trilinears to them,
as well as many OTHER valuable properties that he hasn't yet appreciated.
He will, if he ever looks at them!
John Conway
steve sigur
>
>
>On Tue, 30 Jun 1998, steve sigur wrote:
>
>> I have learned much from John Conway's discussion of coordinate systems
>> natural to triangles, but some reasons that I like Trilinear Coordinates
>> were not mentioned. Although I am trying to use barycentric coordinates
>> as much as possible, I still find myself going back to Trilinears for
>> reasons that are not on Conway's list.
>
> I'm very puzzled by this comment, because the message was all about
>trilinear coordinates. It's obvious in many ways that Steve didn't
>learn enough!
I have no doubt that I have not learned enough. Geometry is a big
subject. This is my way of trying to learn.
>> First
>> Duality (this is related to the algebraic conjugacy discussed below):
>> With Trilinear coordinates for points come trilinear equations for lines.
>> I know there are barycentric lines as well, but trilinear lines can be
>> written in a form that gives them a dual role computing distances to
>> points. I find this very powerful. In fact I prefer to do geometry
>> positing lines first, rather than points first. It seems to me that
>> trilinear lines are better.
>
> All of this applies equally well to all systems of trilinear coordinates,
>not just the orthogonal trilinears that Steve likes. So it gives no reason
>for choosing either of the two major systems over the other.
>
It seems to me that the dual relation between points and lines around a
triangle is well expressed by all coordinate systems. But with trilinear
lines, I can compute in a way I do not think possible with barycentric
lines. Here is an example proof.
Let alpha = 0 be the line corresponding to side a in a triangle. alpha
has the property that alpha(P) is zero if P is on the line and is the
distance from P to line alpha if P is not on the line. This is the
property of trilinear lines that I find valuable.
To prove: if a line goes through the centriod G of the triangle, then Da
= Db + Dc where Da is the distance to vertex A of the triangle (where B
and C are on the opposite side of the line to A).
The line through G can be written L = m alpha + n beta + p gamma = 0,
where m, n, and p are numbers depending on triangle varibles. This first
step is valid for both BA and OT coordinates.The second step cannot.
Consider L(G) = m alpha(G) + n beta(G) + p gamma(G) = 0. Since alpha(G),
beta(G), and gamma(G) are the distances to the sides, they are the
trilinear coordinates, which are, up to a factor, (1/a, 1/b, 1/c). Hence
we have m/a + n/b + p/c = 0 [equation 1].
Now consider L(A). At point A, beta = gamma = 0. Hence L(A) = m alpha(A).
Since alpha(A) is the distance from A to side alpha, alpha(A) = ha, the
altitude to A. Also since L(A) = Da, we have Da = m ha, giving m = Da/
ha. Similarly for n and p (but they are negative since the distances are
taken in the opposite sense).
Substituting into equation 1, we have Da/ (a ha) - Db / (b hb) - Dc / (c
hc) = 0. The denominators are equal hence Da = Db + Dc.
It is to be noted that if we had used the trilinear coordinates for any
other point (say, the orthocenter), we would have gotten a similar
relation for the distances from a line through that point to the vertices
of the triangle.
It is here that I need wisdom. I think this method of changing lines to
distances is unique to trilinears, not shared by barycentrics. Perhaps
barycentric lines can give relations involving areas but I have not been
able to work that out.
---------
>
>> John Conway wrote
>>
>> > 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.
>
> [Then I went on to mistype a formula - I'll correct it now : the
>a-conjugate of ( X : Y : Z ) is
>
> ( X(-a,b,c) : Y(-a,b,c) : Z(-a,b,c).
>
I am glad because those points I _really_ did not understand. I have been
thinking a lot about these "companion" points. But I have one more
confusion which is stated at the end.
> Steve described me as "admonishing" people to use barycentric
>trilinears rather than orthogonal trilinears. ...I AM warning you that if you
use orthogonal >trilinears you'll be missing out on lots of things that
are obvious in barycentric >>trilinears.
>
I am trying to see and understand these things.
> I starred the last line of Steve's letter, because it illustrates
>this and also involves another misunderstanding. Steve points out
>have certain points geometrically related to the point whose orthogonal
>trilinear coordinates are [x:y:z] (his 1/m, 1/n, 1/p) have coordinates
>[-x:y:z], [x:-y:z], [x:y:-z]. This is just as true in barycentric
>trilinears; if the first point is (X:Y:Z) the other three are
>(-X:Y:Z), (X:-Y:Z), (X:Y:-Z).
>
> But he then goes on to say that these are related to [x:y:z] in
>the same way that the excenters are algebraically conjugate to the
>incenter. But that's not what I was talking about. In barycentric
>coordinates, the incenter and excenters are
>
> (a:b:c) (-a:b:c) (a:-b:c) (a:b:-c)
>
>and they are indeed related by changing the signs of the coordinates.
>Since this is true in each system, it gives no reason to prefer one
>to the other.
>
> The algebraic conjugacy I was talking about is much more valuable.
This is what I do not understand. I took your conjugate points as point
strictly analogous to the excenters. I think I understand that structure
well. If cevians meet at P, then the harmonic conjugate lines meet at P1,
P2, P3. But I take it that your algebraic conjugates are not these
points, rather they are more general.
>These four points are related by changing the signs of a,b,c, as
>well! The Nagel point is the point of concurrence of the lines from
>the vertices to the contact points of the incircle, and from its
>barycentric coordinates are (-a+b+c:a-b+c:a+b-c) we immediately
>deduce that the lines form the vertices to the contact-points of the
>a-excenter meet at (a+b+c:-a-b+c:-a+b-c) - just change the sign of a!
>
> This exemplifies my whole point. Steve understands the relationship
>between [x:y:z] and [-x:y:z] (etc), ...What's more important is that he
DIDN'T understand
>the relationship between (X:Y:Z) and ( X(-a,b,c):Y(-a,b,c):Z(-a:b:c) )
>which is just as useful, and which he WON'T ever see unless he switches
>to barycentric trilinears.
>
Yes I do not understand this, but right now I can see those points as
nothing other than more points, not the same as the points I know. I will
play with the Nagel companion points and see if I can make sense of them.
If you (or anyone else) has a good example situation to play with I would
appreciate a suggestion.
Steve Sigur
Kirby Urner
In posting to geometry_research:
> Subject: Re: Conway on Trilinear vs Barycentric
> Author: steve sigur <ssi...@netdepot.com>
> Date: Thu, 2 Jul 98 10:59:10 -0400
Steve wrote:
> I do not think your system is either. Neither trilinears
> nor barycentric coordinates represent distances from a
> common origin. Yours looks like the regular basis vectors
> in 3D space. If yours is in 2D space then I think it is
> different from the other two.
This made sense to me at the time. From what I read, a
hallmark of the barycentrics is the coordinates add up
to some constant. Also they seem to always be used for
planar applications (which is why the link to "trilinear"
coordinates) -- although no doubt a 4-tuple version inside
a tetrahedron has been formalized (quadrilinear coordinates?).
I recall seeing tetrahedral graphs in a chemistry book on
the Gibbs Phase Rule... barycentrics?
With barycentrics, positive coordinates keep you inside the
region defined by the 3 (or 4) vertices, while negatives get
you outside. The idea is a "roving center of gravity"
controlled by "weighted vertices" (redistributing their
constant, sum-total weight).
But 4-tuple quadrays, with 4 basis rays from the common
origin (0,0,0,0) seem more NeoCartesian to me, a completely
analogous system operating in 4 quadrants instead of 8
octants. In other words, the 6 XYZ vectors, 3 basis +
3 not-basis, carve volume into 8 regions, with 8 permutations
of + and - (+++, +-+ +-- ... ---) giving a 3-tuple its "octant
address". In quadrays, the 4 basis vectors carve volume
into 4 regions, and you know which quadrant your point is
in based on what coordinates are non-zero (no negative
signs in the normalized form for any point address).
But there's no stipulation of a "balanced constant weight"
(constant n-tuple sum) or "bounding vertices" with an "inside"
versus an "outside" related to signage in quadrays (at least
not obviously).
Nevertheless, Brian Scott and Robin Chapman, both very
knowledgable mathematicians active on sci.math, have
handed down their decision that quadrays are a subclass
of barycentric coordinates. I mentioned that maybe some
on geometry_research might have a differing opinion, but
Chapman especially was dismissive of geometry_research
posters (me in particular):
I gave up reading geometry.research ages ago
because of its dire signal/noise ratio, mostly
due to Urner's logorrhoeic drivel. The group
should be retitled geometry.remedial. Were I
still reading it I would certainly assent to
Brian's assertion that Urner's "quadrays" are
basically barycentrics.
Robin Chapman (7/31/98)
These guys clearly out-rank me when it comes to having strong
academic credentials in math. Given how the game is played,
I think I have to defer to their judgement. So now my plan
is to write to webmasters who have material relating to
barycentrics on the web and suggest they link to my 'Quadray
Papers' (at http://www.teleport.com/~pdx4d/quadrays.html) as
examplary of how barycentrics have been used to define
a system for rendering a concentric hierarchy of nested
wire-frame polyhedra in object-oriented xBase.
Kirby
Brian M. Scott
Chapman, both very knowledgable mathematicians active on sci.math,
have handed down their decision that quadrays are a subclass
Prof. Chapman will have to speak for himself, but this is an
inaccurate paraphrase of my actual statement that quadrays are
essentially just barycentric coordinates. Since they obviously differ
in detail - including the one that gives barycentric coordinates their
name! - they are certainly not a subclass thereof.
Brian M. Scott
Kirby Urner
>Prof. Chapman will have to speak for himself, but this is an
>inaccurate paraphrase of my actual statement that quadrays are
>essentially just barycentric coordinates. Since they obviously
>differ in detail - including the one that gives barycentric
>coordinates their name! - they are certainly not a subclass
>thereof.
>
>Brian M. Scott
Hmmmm. I find this somewhat confusing Brian. "Essentially
barycentric" but "differ in detail - including the one that
gives barycentric coordinates their name!... not a subclass
thereof." Some might define this "detail" to be "essential"
given it sounds like nomenclature is at stake -- i.e. do we
call them "essentially barycentric" if differing in some
detail that essentially defines "barycentric"?
To get more specific, here's an excerpt from a draft of an
article I'm writing for a professionals' magazine (my signed
Writer's Agreement keeps me from disclosing which one).
I post it here for peer review -- not wanting to mislead
my readers:
If this all seems pretty exotic, nothing like what you
remember from high school geometry class, it is. To the best
of my knowledge quadrays have only been on the scene for
about 20 years, were invented by a number of individuals
working both collaboratively and solo (e.g. David Chako,
D. Lloyd Jarmusch, Josef Hasslberger and myself). [SNIP]
Why use quadrays? Note the simple, whole number coordinates
I get for my tetrahedron (above). Many other shapes come out
with similarly simple data -- reason enough for a mathematician
to toy with this gizmo. And given the ideas are familiar
enough to high school students with some background in
Cartesian coordinates and vector addition (putting arrows
tip-to-tail), why not clue them in as well? The ideas relate
back to more conventional topics, while keeping minds flexible,
reminding kids (and their teachers) that xyz isn't the only
game in town.
Comments welcome by private email: pd...@teleport.com (or
post here if so moved). Should I be telling my readers anything
about barycentric coordinates i.e. mentioning their inventors,
talking about how quadrays are "essentially the same thing"?
For obvious reasons I don't want to just lift Dr. Chapman's
"barycentrics made difficult" characterization, as the whole
point of my article is how easy and accessible-to-kids is
this NeoCartesian game.
Kirby
PS: as to the sci.math roots of this query (re how to taxonomize
quadrays among many things -- and I still prefer "NeoCartesian" to
"essentially barycentric") here's an excerpt from Dr. Chapman's
post to that newsgroup of July 29, 1998, full text archived courtesy
of Deja News.
Major reference:
http://www.teleport.com/~pdx4d/quadrays.html
=====
Kirby Urner (quoted by RC):
Seems like sometimes mathematicians give lip service to how
consistency and precision are the hallmarks of their
game-playing, but then turn up their noses if someone comes
along with a consistent and precise symbol game that just
doesn't happen to be the same as theirs.
Robin Chapman (commenting):
Urner's quadray system (or barycentrics made difficult) is
easily translated to Cartesians and vice versa. The only
difference is that Cartesians are easier to use. The easiest
way of doing quadray computations is to translate to
Cartesians, do the computations in Cartesians, and translate
back!
Kirby Urner (quoted by RC):
...not essentially barycentric, I disagree...
Brian Scott (quoted by RC, responding to KU):
Eh? Of course they are: the basic idea's exactly the same.
Robin Chapman (responding to BS):
Well said, Brian.
=====
Kirby Urner
My article touching on quadrays in the context of 'Teaching
Object-Oriented Programming with Visual FoxPro' is now in
circulation, in the March 1999 issue of FoxPro Advisor, The
Developer's Guide to Microsoft Visual FoxPro and FoxPro 2.x,
ADVISOR MEDIA Inc., starting on page 48.
The primary web page reference is to:
http://www.teleport.com/~pdx4d/quadrays.html
-- part of my Oregon Curriculum Network website (OCN).
This page links to numerous related papers and downloadable
source code, including to a Java implementation which includes
code for converting back and forth among xyz, polar and quadray
coordinates.
For those unfamiliar with quadrays, this Neo-Cartesian system
uses 4-tuples instead of the familiar xyz 3-tuples, labeling
the origin as (0,0,0,0) and the 4 basis vectors to the vertices
of a regular tetrahedron as (1,0,0,0)(0,1,0,0)(0,0,1,0) and
(0,0,0,1).
These four vectors span volume (given vector addition and
scalar multiplication of the usual sort), giving a unique,
simplest-terms 4-tuple (a,b,c,d) for each point, with
at least one term equal to zero and the rest positive.[1]
(n,n,n,n) is an additive identity i.e. Q + (n,n,n,n) = Q --
because (n,n,n,n) simply represents four equal vectors to
the corners of a regular tetrahedron with a zerovector net.
Given all basis rays are positive, no simplest-terms 4-tuple
includes any negatives, although the negation operator does
have a clear meaning (same as in xyz -- changes a vector's
orientation by 180 degrees). For example: -(1,1,0,0) =
(-1,-1,0,0) = (-1,-1,0,0) + (1,1,1,1) [identity] = (0,0,1,1).
My implementation of quadrays nests the "home base" tetrahedron
in a unit-radius spheres packing context (fcc) and also defines
this tetrahedron to be the unit of volume.
The home base tetrahedron and its self-dual, a tetrahedron with
quadray coordinates (0,1,1,1)(1,0,1,1)(1,1,0,1)(1,1,1,0) -- the
negatives of the four above -- intersect to define the cube of
volume 3.
This cube's dual, an octahedron with coordinates (1,1,0,0)
(1,0,1,0)(1,0,0,1)(0,1,1,0)(0,1,0,1)(0,0,1,1), has a volume
of 4.
The rhombic dodecahedron, with cube + octahedron vertices,
has a volume of 6, and is a space-filler within which the
unit-radius fcc spheres nest and "kiss" through the face
centers (12 K-points where long and short diagonals of the
12 rhombic faces intersect, and where long diagonal is equal
in length to the inter-spheric center-to-center prime vector
of 1 interval = 2 radii).
The cuboctahedron with quadray coordinates {2,1,1,0} -- where
the curly braces signify "all 12 permutations of the enclosed
terms" -- has a volume of 20.[2]
The above cited article mentions all of this in passing, but
is primarily about how to use Visual FoxPro in conjunction
with freeware off the internet to (a) render colorful geometric
shapes of intrinsic interest to students (including geodesic
spheres -- website) and (b) learn object oriented programming
concepts in the process.
Those of you who have explored the interface between 20th
century philosophy and geometry will recognize the above
approach to volumetric mensuration, with the tetrahedron
as primary, as essentially the same as the one pioneered
by Dr. R. B. Fuller in his '4D geometry' -- wherein '4D'
refers neither to "3D + Time," nor to Cartesian hyperspace,
but to the tetrahedron's paradigmatic status as the minimal
volumetric system.[3]
In Fuller's '4D geometry', all objects are ab initio volumetric,
even if we restrict their degrees of freedom to a plane or a
point, i.e. no 'dimension ladder' with rungs labeled 0,1,2,3
(with fractional dimension interpolations) is natively defined
in this particular philosophical language.
Traditional classroom geometry ports to this alternative
definitional environment with only minor conceptual and
terminological adjustments i.e. Euclidean and Cartesian language
games co-exist with 4D geometry without significant friction
in this new curriculum context.
The quadrays apparatus has been developed by individuals
working solo, at first unbeknownst to one another, but coming
together via the internet to contribute various puzzle pieces.
The article identifies some of the key players in passing, with
more details at my website.[4]
As for myself, I am more an implementor than an inventor of the
quadrays game, and do not push them at my website as any kind of
"next big thing". I'm personally most interested in using quadrays
in low key fashion as one more pedagogical device for assisting
in philosophical investigations into the definitional framework
(cultural) within which our mathematical style of thinking is
embedded, in directions pioneered by the late philosopher-engineer
Ludwig Wittgenstein.[5]
Kirby
[1] although it makes sense to hyperlink quadrays to barycentric
coordinates for pedagogical purposes, quadrays are not a subspecies
of the barycentrics.
Quadrays also include analogous implementations as planar or
linear games i.e.
Linear:
<-|---------*---------|->
(0,1) (0,0) (1,0)
and:
Planar:
(0,0,1)
|
|
|
* (0,0,0)
/ \
/ \
/ \
(1,0,0) (0,1,0)
[2] Robert Gray has proved David Chako's conjecture that any
tetrahedron with fcc vertices will have a whole number volume
relative to the unit-volume tetrahedron comprised of 4 intertangent
unit-radius fcc spheres (i.e. the quadrays "home base" tetrahedron).
It follows that any tetrahedralizable volume with fcc vertices is
likewise whole-number volumed.
[3] '4D geometry' is embedded within Fuller's "explorations in
the geometry of thinking" or "synergetic-energetic geometry"
or "synergetics" for short, a 20th century philosophy originally
published in two volumes, by R. Buckminster Fuller, USA Medal
of Freedom winner, and E.J. Applewhite, former deputy inspector
general of the CIA. This work is now available on the web via
my http://www.teleport.com/~pdx4d/links.html, thanks to Robert
Gray et al.
[4] cite http://www.teleport.com/~pdx4d/quadintro.html
[5] cite http://www.teleport.com/~pdx4d/quadphil.html
IXAB...@aol.com
from each coordinate and you get (1,0,0,-1) and the twelve unique permutations
are the vertices of a cuboctahedron in my system. But the permutations of
(1,0,0,0) in quadrays are not the centers of closest packed spheres. They are
something that requires fractions in my system. The idea is to use integers
(mod a prime), and get only the centers of closest packed spheres.
By the way, is arithmetic remainder (as well as modulo) a prime a field?
Bucky wrote that the tetrahedron's four dimensions refer to the distances from
the midpoint of the tetrahedron to the centers of the four faces. The
advantage of that is to represent any size (positive or negative) tetrahedron
anywhere with four coordinates. There is a way to transform from my system to
quadrays just as there is a way to transform to XYZ. (or if the four
coordinates do not sum to zero, to XYZT).
Cliff Nelson
<A HREF="http://forum.swarthmore.edu/epigone/geometry-research/brydilyum">RBF
electronic notebooks by Clifford J. Nelson</A>
Kirby Urner
>my system. Subtract one from each coordinate and you
>get (1,0,0,-1) and the twelve unique permutations
>are the vertices of a cuboctahedron in my system.
Yes, same number of permutations, given one of the terms
appears twice, other 2 uniquely.
>But the permutations of (1,0,0,0) in quadrays are
>not the centers of closest packed spheres.
Actually, in quadrays {1,0,0,0} (permutations) *are* at the
centers of closest packed spheres. But (0,0,0,0), the origin,
is not -- it's in the hole between the 4 intertangent spheres,
at the center of the regular tetrahedron.
Do you include volume calculations in your system? Makes sense
that if your cubocta has {1,0,0,-1} for coordinates, that your
related tet would have fractional terms, as their volume ratio
is 20:1 (or is it, in your gizmo?).
>They are something that requires fractions in my system.
>The idea is to use integers mod a prime), and get only
>the centers of closest packed spheres.
If you move my (0,0,0,0) to the center of a closest packed sphere,
then you're correct, {1,0,0,0} are in the holes around that sphere.
{1,0,0,0} + {1,1,1,0} = the vertices of the volume 3 "duo-tet"
cube, with vertices in the tetrahedral holes of the fcc (at the
termini of the short diagonals of the space-filling, sphere-
containing, rhombic dodecahedron of volume 6).
So if (0,0,0,0) is aligned with an IVM (or fcc) sphere center,
then all the other sphere centers will be linear combinations
of {2,1,1,0} i.e. add from those 12 at will for a "random turtle
walk" that "connects dots" in the fcc (cuboctahedral) sphere
packing arrangement.
> By the way, is arithmetic remainder (as well as modulo)
> a prime a field?
Not sure I understand the question. Remainders needn't be
primes, merely less than the divisor.
> Bucky wrote that the tetrahedron's four dimensions refer to the
> distances from the midpoint of the tetrahedron to the centers
> of the four faces. The advantage of that is to represent any
> size (positive or negative) tetrahedron anywhere with four
> coordinates.
I essentially agree with you here. Note that if you depict 4
vectors to the vertices of a tetrahedron, you're likewise showing
the same vectors as penetrating the _face centers_ of the dual
tetrahedron (the invert). So quadrays _do_ go through tetra-
hedron face centers. Where they 'stop' doesn't really matter
that much (you can always grow or shrink any tetrahedron).
But yes, Fuller's idea was to begin with volume as the starting
point for conceptualization. Abbott's 'Flatland' neglects the
observer, i.e. if you imagine a line, you also imagine the space
in which that "observer-observed" relationship occurs. Synergetics
brings the camera (which originally meant "room") or the "mind's
eye" back into our thinking.
The separation of relationship, between a point and an observer
of that point, creates volumetric twoness (the twonesses of poles
i.e. axial rotation, and of concave/convex). Experientially, our
awareness of geometric objects is in this context of volume. This
is all philosophical content, at the definitional level. You
have to start somewhere and synergetics starts here, with the
concept of volume, of context as containment -- minimally a
container vs. contained relationship.
The next step is to give a minimal shape to that context, to
signify containment. Just as 3 points define a plane, making
the triangle the minimum signifier of a flat 'two dimensional'
surface (which one observes in a spatial context), so four non-
coplanar points define space, which gives us the 6 lines of
relationship, and the separation of volume into 'internal' and
'external' sets (plus the system-divider itself). Of all
polyhedra, the tetrahedron has the minimum inventory of edges,
vertices and faces. And this becomes the new unit of volumetric
mensuration.
In stressing that conceptual volume is '4D' in Synergetics,
Fuller is deliberately going against the grain of standard
academic usage, according to which we're all drilled. We
all know to think of volume as comprised of three "linearly
independent" dimensions: height, width and depth. In
'Synergetics' you find passages designed to counter this
programming, suggesting that we have no conceptual experience
of volume minus one of its "independent" component dimensions
-- any more than we have an experience of "four mutual perpend-
iculars" (or more), I would add. This is not an empirical
limitation so much as a logical one, or what the later
Wittgenstein would call a "grammatical" aspect of our experience
(language and experience ultimately having no "internal
boundary" to keep them apart: Physus = Logos).
Fuller thereby sets up a kind of "noise" in both directions:
neither subtracting perpendiculars, nor adding them, gets us
away from primitive, tetrahedrally defined volume -- which he
defines as 4D, just to help break the spell of the older paradigm.
He does add further dimensions however. But these are in the
direction of "more reality" i.e. dimensions having to do with
time, energy (more of the "physics meaning" of that key term).
The added dimensions, which differentiate an abstract, purely
imaginary cube, from a physically realized one, come under the
heading of "frequency" in Synergetics.
This attempt to embrace the philosophical distinction between
the Platonic cube and the Empirical experience of special case
cubes, within a consistent language, is partly what makes
'Synergetics' a philosophical work and not a mathematical one.
Mathematicians invoke the "Platonic versus Real World" distinction
in their writings, but don't consider such chatter properly a
part of their formalisms, which are strictly on the "Platonic"
side.
To talk about the "difference" between conceptuality in pure
principle and energized spacetime, is generally not undertaken
as an exercise in mathematics per se -- this "difference"
corresponding to an interdepartmental interface within the
academic context, a boundary between discourse A and discourse B.
But Fuller's commitment was to subsume the multiple academic
languages and anchor them within a consistent gridding system
or conceptual framework -- a kind of old school commitment to
comprehensiveism which we rarely find on the university scene
these days, in any department.
Which isn't to say we can't think along the old lines in whatever
special case shop talks, e.g. we needn't deny ourselves the freedom
to write computer programs wherein nodes and edges have as many
tuples as we like. And we will continue to apply spatial metaphors
to so-called hyperdimensional polytopes. But we might also choose
new metaphors just as well e.g. look at n-tuple encodements of
visualizables (the 3D shapes extractable-by-algorithm from
n-tuple data storage formats) as a kind of "data compression"
(hyperspace as a kind of encrypted "zip" format, wherein methods
such as "rotation" are operationally defined (metaphors consistent
with how n-tuple mathematics is actually used to support cell
phone channel multi-plexing in the engineering department)).
To accept Fuller's language as internally consistent in its own
way is not to break from usage patterns which characterize the
bulk of contemporary mathematics -- but it _does_ serve to help
counter some of the dogmatic hardening of the mental arteries
associated with the latter, symptomatic of the unquestioning
acceptance of various authorities over the centuries i.e. it's
not "God given", nor even "a priori", but "cultural" that we say
"space is three dimensional". An intelligent, well-schooled
layperson might reasonably say otherwise, and not be less sane
therefore.
I am hopeful that a more enlightened, Renaissance era will ensue,
as we learn from the case history surrounding 'Synergetics' and
realize the price of over-specialization, which almost succeeded
in preventing future generations from appreciating the many life-
supportive advantages contributed by this important century
thinker and engineer-philosopher.
> There is a way to transform from my system to
> quadrays just as there is a way to transform to XYZ. (or if the four
> coordinates do not sum to zero, to XYZT).
I don't doubt it.
Kirby
Clifford J. Nelson
Kirby Urner wrote:
<< Not sure I understand the question. Remainders needn't be
primes, merely less than the divisor. >>
Is arithmetic, remainder (as well as modulo) a prime, a field? The
remainder function can have a sign of positive or negative, but the
mod function is always positive.
Cliff Nelson