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Topic: Conway on Trilinear vs Barycentric coordinates
Replies: 13   Last Post: Feb 16, 1999 8:05 AM

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steve sigur

Posts: 139
Registered: 12/6/04
RE: Conway on Trilinear vs Barycentric coordinates
Posted: Jul 3, 1998 4:54 PM
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>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.

>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)
> 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

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