
Re: Random Triangle Problem
Posted:
Jul 30, 1997 4:24 AM


First a few comments on your comments.
I am not a newbie. I did not start this post series. I have shown all my working, including correcting my errors. I have read most contributions, all with respect. I have not insulted, sneered or abused anyone in this post or assumed anything about them other than that they appear to be interested in actually solving the problem as stated,
which is more than can be said for yourself. Your abuse does not further any argument you might have made, on the contrary, it may show that others understand your argument better than you do yourself.
I will finish by including my final comments, which I hope make clear the differences between us (me and others) on the 'random triangle' problem. Please read it if you really wish to understand someone elses approach, rather than just looking for confirmation/agreement with your own prejudices, without going into the details (which may prove uncomfortable for your position). The bottom line ssurely, should be to come out of this with at least a better understanding of the problem, and at best a solution.
tony richards <tony.richards@rl.ac.uk> wrote: >IMHO, it would be useful to restate the problem as originally posed. (not by me I hasten to add) >I believe the first poster mentioned ' probability of OBTUSENESS given >three randomly chosen POINTS'. > >An early reply stated, without proof, that the result obtained for >three randomly chosen points 'must be' equivalent to >triangles formed by three randomly chosen LINES. > >IMHO, the two situations are NOT equivalent, and so need not necessarily produce the same >answer. I do however accept that three randomly chosen LINES will give an obtuse triangle >75 percent of the time. > >The reason I believe that the two processes for generating random triangles are not equivalent >is as follows. > >Given two randomly chosen points, you have defined TWO VERTECES but only ONE LINE, whose >direction can be regarded as randomly selected in direction space. However, as soon as the >third POINT (or VERTEX) is selected, you SIMULTANEOUSLY FIX the other two LINES or sides, >and these CANNOT be randomly related to each other in direction space. > >OTOH, >Given two randomly chosen LINES in direction space, you have defined only ONE VERTEX, whose >location can be regarded as randomly selected in the plane. However, as soon as the third LINE >is randomly selected, you SIMULTANEOUSLY FIX the other two VERTEXES (or points), and the >locations of these CANNOT be randomly related to each other. > >Therefore, IMHO, if you actually model the process using random POINTS, you may well get a >different answer than you would get when modelling random intersecting LINES. > In answer to another polite reply I posted the following
>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 proportion (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 'but I don't slag people off ' UK 'unless sorely provoked. Proving that' 'my reasoning is unsound does not count' 'as provocation in a newsgroup. Abuse does'

