From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Triangulation Problem
Date: 27 Mar 1995 18:42:52 GMT
In article <3kouh2$q4l@girtab.usc.edu>,
Ash Morgan wrote:
> Knowing the the length of each side of the triangle A,B,and C
> and the angles of the triangle opposite each side a,b,c
> and the three internal angles formed by the point inside the
> triangle each of the vertices of the triangle: P, Q, R
>
> Is is possible, and if so how, to determine the x,y coordinates of
> the point relative to a coordinate system with the origin located
> aht the lower left vertex and the x-axis parallel to side C.
If you want to work the whole thing out in coordinates, you have a
triangle with vertices at (0,0), (C,0), and (D,E), where
D=(C^2-B^2+A^2)/(2C) and E = sqrt(A^2-D^2). If you draw segments from
the vertices to an interior point (x,y), you form angles which can be
determined by taking dot products. For example, the angle you call R
has cosine ( (0-x)*(C-x) + (0-y)*(C-y) )/ sqrt(x^2+y^2)*sqrt((x-C)^2+y^2).
So if the three angles are given, you get three equations to determine
x and y. The one above, for example, may be written
(x^2+y^2)((x-C)^2+y^2)=cos^2(R) (x^2-Cx+y^2-Cy)^2; the others, also, are
quartics. (Actually the three equations are redundant, assuming that
the angles P, Q, and R do indeed sum to pi). Thus the point (x,y)
can be found by intersecting the quartic curves. Numerically, this can
be done by eliminating y, say, to get a polynomial of 16 in x, which
can then be solved by, say, Newton's method. (I asked Maple to eliminate
y from the two equations symbolically; the result scrolled through many
screenfuls of data which I won't recreate).
I would guess that the construction can be done with ruler and compass,
that is, that the equation can be solved with a succession of extractions
of square roots; but I didn't see an elegant construction.
dave
==============================================================================
[Maple input file for creating the quartic referred to in the previous
"triangulation" correspondence. (Originally dated 3/29/95) -- djr]
readlib(eliminate);
eq1:=(x^2 - C*x + y^2)^2 = (x^2 + y^2)*((x-C)^2 + y^2)*(C1);
eq2:=(x^2 - D*x + y^2 - E*x)^2 = (x^2 + y^2)*((x-D)^2 + (y-E)^2)*(C2);
ans:=eliminate({eq1,eq2},y);
save ans, `temp`:
#remark(`temp` now has ans:=[{y=blah},{x*yuk}]; where yuk=resultant condition);
#remark(C1=cos(P)^2, C2=cos(R)^2 (I think));
#Output file is about 55K...
==============================================================================
From: Ash Morgan
Subject: Triangulation Problem
To: rusin@washington.math.niu.edu
Date: Tue, 28 Mar 1995 11:50:15 -0800 (PST)
Hi!
Thanks for your input on my triangulation problem. I had tried solving it
using line intersections and it just wasn't quite working right. I should
have just used vectors in the first place. Anyway, we don't have Maple at
USC, so I've been trying to use Mathematica to eliminate y from the
equations but it's been running for several hours and hasn't been able to
do it. If you would could you eliminate y from the equations using Maple
and e-mail me the results. Thanks a lot.
(x^2 - Cx + y^2)^2 == (x^2 + y^2)((x-C)^2 + y^2)(Cos[R])^2
(x^2 - Dx + y^2 - Ex)^2 == (x^2 + y^2)((x-D)^2 + (y-E)^2)(Cos[P])^2
| |
| |
Ash
\ /
`.__.'
==============================================================================
Date: Wed, 29 Mar 95 12:13:26 CST
From: rusin (Dave Rusin)
To: ashleymo@scf.usc.edu
Subject: Re: Triangulation Problem
You know, I think there has to be a geometric (ruler and compass) solution
to your problem. I took that terrible quartic I sent you yesterday and
tried specializing the values. I chose several sets of values which appear
to be unrelated (e.g. C=71, D=45, E=103; C1 (=cos(R)^2) = 4/9, C2=2/17.)
This gives a polynomial with rational coefficients. I then asked Maple to
determine the galois group of the quartic over Q. In every case I tried,
I either got that the quartic factored into two quadratics, or the quartic
was irreducible with galois group D4 (the dihedral group of order 8). In
the latter case, this means there is a sequence of subfields between the
rationals and the field generated by the roots of the quartic, such that
each subfield is a quadratic extension of the previous one, i.e., it is
generated from the previous one by adjoining some square root.
Well, it's basic geometry that this is precisely the condition that the
point in question be constructible by a sucession of drawing lines and
circles thru existing points and then looking for points of intersection.
So, at least in all the cases I tried, this means the point you seek can
be found by such a construction. I am loathe to actually attempt the
construction until I can be sure there is a method which will work in
all cases. That is, I want to know the galois group is _always_ D4.
I also tried to figure out why there are 4 values for x. In the examples
I tried, all the roots are real. I think what happens is that the polynomial
finds the x-coordinates knowing not the angles P Q and R but rather
the squares of their cosines. Thus, if you set the same problem with
P replaced by pi-P or Q replaced by pi-Q, you'll have to solve the same
equation. Thus the x-coordinates for all 4 of the points must be the
solutions to the polynomial. It may be that the use of the third equation
will eliminate some of the possibilities (but I doubt it).
I don't know what use you intend to put this to. Certainly if you only need
to find the point numerically once the original triangle is given, you only
have to solve some numerical quartic polynomials, which is a piece of cake.
But I still think it's an interesting question to find a geometric construction.
dave