From: m9305474@student.anu.edu.au (John McLaughlin)
Newsgroups: sci.math
Subject: obtuse geometry question
Date: Sat, 29 Apr 1995 11:16:23 +1000

Hi,
  I need to solve a triangle problem in obtuse geometry (for a program
algorithm). It is as follows:


                            p2
                           /  \
                        b /    \ a
                         /      \
                        /        \
                       /          \
                      p1----------p3
                             c

p1, p2 and p3 lie on the unit sphere.
The angles between the points are a,b and c as indicated.
The problem is, given a, b, c, p1 and p2 how do I find
p3 = ( 1, theta3, phi3 ) (in spherical coords) ?
Presumably there is some quadratic solution giving two values
for p3. That's ok.

Any email replies would really be appreciated. I will post the 
solution.

-John McLaughlin

==============================================================================

From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: obtuse geometry question
Date: 4 May 1995 15:16:12 GMT

In article <m9305474-2904951116230001@150.203.111.143>,
John McLaughlin <m9305474@student.anu.edu.au> wrote:
>                            p2
>                           /  \
>                        b /    \ a
>                         /      \
>                        /        \
>                       /          \
>                      p1----------p3
>                             c
>
>p1, p2 and p3 lie on the unit sphere.
>The angles between the points are a,b and c as indicated.
>The problem is, given a, b, c, p1 and p2 how do I find
>p3 = ( 1, theta3, phi3 ) (in spherical coords) ?

Let's use an orthonormal basis in which we write
	p1=e1
	p2=z e1 + y e2
	p3=w e1 + v e2 + u e3
Then the conditions given imply (using dot products)
	cos(b)=z
	cos(c)=w
	cos(a)=zw+yv
and since the  p_i  are on the unit sphere, we have lengths=1:
	z^2+y^2=1
	w^2+v^2+u^2=1
That's five equations in 5 unknowns, whose solution is
	z=cos(b)	
	y=+-sin(b) [we can replace  e2  by  -e2  to assume y=sin(b)]
	w=cos(c)
	v=[cos(a)-cos(b)cos(c)]/sin(b)
	u=+-sqrt(sin^2(c)-v^2) 
Actually, you could use these to _define_ e2 as  (p2 - cos(b) p1)/sin(b).
Then you can compute  e3 = e1 X e2  (or its negative, e2 X e1; different
choices will lead to the two possible points  p3  meeting the conditions
of your problem -- the ones on either side of the arc  p1 p2).

So one solution to your problem is to compute w, v, u as above, and
compute  e2 and  e3  as indicated, then deduce 	p3=w e1 + v e2 + u e3
(computed in cartesian coordinates). If I've made the right substitutions,
this makes  p3  be  1/[sin(b)]^2  times the following vector:
	  [cos(c)-cos(b)cos(a)] p1 +
	+ [cos(a)-cos(b)cos(c)] p2 +
	+-sqrt[1-cos^2(a)-cos^2(b)-cos^2(c) + 2cos(a)cos(b)cos(c)] p1 X p2
You can then convert to spherical coordinates if you wish.

dave