
Re: Random Triangle Problem
Posted:
Jul 29, 1997 4:04 AM


The reason I persist with this problem, is that I remain unconvinced by handwaving arguments and special pleading involving 'common sense' assumptions which turn out to be unwarranted. Several contributions to this subject have stumbled because of this.
I have spotted the one remaining fault with my previous effort, which no one spotted, but which I now correct here.
I obtained the following expression for the probability for an OBTUSE triangle
P(OBTUSE)=[ 1  abs(x)*2*L/(4*L^2) +(pi*abs(x)^2/4)/(4*L^2)]*P(x)
where the first point is located at the origin (or vice versa) of an infinite plane, side 2*L in X and Y, the second point is located a distance x from the first and the third is located in (a) the area to the left of X=0, or in (b) the area to the right of X=x, or in (c) the circle diameter x centred at X=x/2 .
I then made the erroneous assumption that the probability/weighting function for the second point to be at distance x from the first point is P(x)=dx/(2*L) when further thought shows that it must actually be P(x)=2*pi*x*dx/(pi*L^2). That is,the probability that the second point is distance x away from the first is actually proportional to the area of an annulus, radius x, thickness dx, centred at the first point, divded by the total area available (pi*L^2) (orienting the axes after the location of the second point so that the second point is along X does not mean that the spacing is only along the x axis)
With this correct weighting/probability function, the result for the probability integrated over x, taken over the range 0<x<L, is
P(OBTUSE)= 1 (1/3) + pi/32 = 0.76484.
This is close to, but not identical to the 3/4 everyone seems to expect.
My final question is , 'why persist in believing that the probabilities for each vertex being obtuse must be the same', when it is clear that, given the first two points, the probabilities for the third point falling within the three different areas of the infinite plane which will result in the triangle being OBTUSE are clearly different?'.
 Tony Richards 'I think, therefore I think I am right this time' Rutherford Appleton Lab ' UK '

