
Re: Random Triangle Problem
Posted:
Jul 24, 1997 10:45 AM


In article <5r549u$p8u@newton.cc.rl.ac.uk>, tony richards <tony.richards@rl.ac.uk> writes: > 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.
Let the infinite unoriented lines which form the triangle be L1, L2, L3. Orient L1 by choosing arbitrarily one of the two possible ways of marking an arrow on it, and make this the positive direction of the xaxis. The orientation of the yaxis is now determined.
Since the lines form a triangle, neither L2 nor L3 is parallel to the xaxis. So for each of these two lines, there is precisely one way of marking an arrow on it along which y increases. Mark these arrows. We have now oriented all three lines.
The anticlockwise angles t2 from L1 to L2, and t3 from L1 to L3, are both between zero and +180 degrees. If you will accept that t2 and t3 are independently and uniformly distributed on this interval, then the 3/4 result follows.
For in all cases (I think) the angles of the triangle are min(t2,t3), 180  max(t2,t3), max(t2,t3)  min(t2,t3)
and the events that each of these is > 90 are mutually exclusive so that their probabilities can be added.
P[min(t2,t3) > 90] = P [t2>90 & t3>90] = 1/4
P[180max(t2,t3) > 90] = P[max(t2,t3) < 90] = P[t2<90 & t3>90] = 1/4
To get P[max(t2,t3)min(t2,t3) > 90] one can do an integration, or draw the square 0<= t2/180, t3/180 <= 1 and shade in two triangles each of area (1/2)(1/2)(1/2) to get the answer 1/4.
So by this method the probability of obtuseness at each angle is 1/4 and the overall probability 3/4.
 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)

