
Re: Random Triangle Problem
Posted:
Aug 1, 1997 8:19 AM


Robert Hill (eclrh@sun.leeds.ac.uk) wrote: : (4) x and y coordinates of each vertex independent normal N(0,1) : (if my memory is correct this distribution is isotropic, : i.e. for any fixed a it makes x cos a + y sin a also N(0,1)): : 249791 249989 249974 250246
: Without running a statistical test, the results for (4) looked as though : they might be compatible with my vague intuition that the theoretical : answer for this case may be exactly 3/4.
It is. Angle BAC is obtuse iff A is closer to the midpoint of BC than B and C are; that is, iff the vector A  (B+C)/2 is shorter than the vector (BC)/2. By the wonders of the normal distribution, if A, B, C are independent with the same circularly symmetric 2dimensional normal distribution [is this the right terminology?], then A  (B+C)/2 and (BC)/2 are independent with circularly symmetric 2dimensional normal distributions, the former having 3 times the variance of the latter. A little calculus then shows that the probability that the former has smaller magnitude is 1/4. (Is there any easy way to see that this must be true without using calculus? If "3" is replaced with "a", the probability becomes 1/(a+1).)
 John Rickard

