
Re: Random Triangle Problem
Posted:
Jul 24, 1997 11:05 AM


In article <33D6B081.35ED@navpoint.com>, "T. Sheridan" <stopjunkmailsparky@navpoint.com> writes: > I think the problems is you need to know if the region is bounded. > If it's on an infinite plane then the probability of the triangle being > obtuse aproaches 1. > > The reason is simple. Just imagine two points are picked. (A,B) They > always form a line. [AB} Now there are two perpendiculars extending > from each point. > . * . >   >   > *  *  * >   > AB >   >  *  >   > . *  > > > They define two half planes wherein the third point will create an > obtuse. That's virtually all the area. Within the two perpendiculars > There is a small band of area close th ethe line where the triangle can > also be obtuse but the rest is acute. > > If the region is bounded then the calculs gets miserable. > > Unless I'm just wrong :)
The "small band" is circular.
As I've pointed out in reply to Tony Richards in this thread, all variants of this approach make an obtuse angle at the third vertex less likely than at the other two, and are therefore presumably wrong.
I think that your version (which is a bit less sophisticated than his) is a disguised form of the following old paradox: two people each choose a positive integer "at random" (with an assumed uniform distribution). Whatever the first person's number, almost all possibilities for the second person's number are larger, therefore with probability one the second person's number is larger. But you could say the same for the first person's number.
 Robert Hill
University Computing Service, Leeds University, England
"Though all my wares be trash, the heart is true."  John Dowland, Fine Knacks for Ladies (1600)

