> > >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.