
Re: Random Triangle Problem
Posted:
Jul 23, 1997 10:26 AM


mathwft@math.canterbury.ac.nz (Bill Taylor) wrote: >Mike Housky <mike@webworldinc.com> writes: >> >> > Derive a method of finding the chances of 3 random points placed on >> > a plane that will yield an obtuse triangle. > >Any answers to openended problems of this type are bound to be somewhat >subjective, or a matter of taste. > >But in this case, the answer seems fairly clear (and has appeared on Usenet >many times before). > >Most folk would consider a triangle formed by "3 randomly chosen points" to >be the same as one formed by "3 randomly chosen lines", and this second >formulation, though tougherlooking at first glance is in fact much cleaner. > >The three lines, whatever else they have, should have orientations that are >independent and uniform in (0 , 2pi) ; and this means choosing three numbers >at random in that range. Once done, we then see that their actual placements >ARE IRRELEVANT TO THE SOLUTION ! (Merely replacing triangles by similar ones.) > >This last observation leads me to suggest that this is a "complete" solution >to the problem, and disputes of taste and subjectivity essentially decided. > >Thus, the problem is fairly easily solved. Let one line be the xaxis; if the >other two have the same signed slope (prob 1/2), an obtuse is certain. If they >have opposite slopes there are equal probs of obtuse or acute, (as the angle >between then can be on either side of 90 deg).
IMHO, this is where you make an invalid assumption. The third line is not free to make ANY angle between 0 and 180 degrees with the second line, since it is constrained to have a slope opposite in sign to the second line, WHEN MEASURED RELATIVE TO THE FIRST LINE. You therefore have a skewed distribution for the third line, given the first and the second lines, so your assumption of equal probabilities of >90 and <90 degrees between the second and third lines is unwarranted. the answer therefore cannot be 0.75, if that is what your argument gives since it has the above flaw. > >So putting together, this gives... P(obtuse) = 3/4 . > ~~~~~~~~~~~~~~~
 Tony Richards 'I think, therefore I am confused' Rutherford Appleton Lab ' UK '

