
Re: Random Triangle Problem
Posted:
Jul 30, 1997 4:00 AM


kramsay@aol.com (KRamsay) wrote: >In article <5rk86u$np6@newton.cc.rl.ac.uk>, tony richards ><tony.richards@rl.ac.uk> writes: > >My final question is , 'why persist in believing that the probabilities >for each >vertex being obtuse must be the same', when it is clear that, given the >first >two points, > >What do you mean by "first two"? One is picking a triangle at random. > I mean that before you can 'pick a triangle' you first have to construct it and the original post referred to the method of construction involving 'choosing' three points 'at random' on a (presumably flat) surface, the points being made the verteces of a triangle. So I was referring to the first two of the triple of points making up the verteces of a triangle.
My argument is and remains, generating triangles by joining up three randomly generated points (over an infinite plane) will produce a slightly different proportion of obtuse angles triangles (0.7648414 , which = 1 (1/3)+pi/32) than the proprtion (0.75) which will be produced by generating triangles as the included space between three randomly generated straight lines (random in direction and location on the plane).
 Tony Richards 'I think, therefore I am confused' Rutherford Appleton Lab ' UK '

