
Re: Random Triangle Problem
Posted:
Jul 23, 1997 10:07 AM


eclrh@sun.leeds.ac.uk (Robert Hill) wrote: >In article <5r2l8b$eac@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. >> > >> without loss of generality, we can assume point 1 is at the origin with probability =1 >> so P1(anywhere)=1.0. (snip)
OK. I made a mistake. This one looks better though.
Without loss of generality, let the first point be the origin of an infinite plane, the plane being 2*L along x and 2*L along y, with total area Ap=4*L^2.(As long as L is very large, then the answer turns out not to depend on L)
Let the second point be on the X axis at x. The area of the half plane opposite to the halfplane containing x = H = 2*L^2 The area of the strip between 0 and x is S = abs(x)*2*L The area of the rest of the half plane beyond x = (Labs(x))*2*L Let the area of the circle, diameter x be Acirc= pi*abs(x)^2/4 Let the area of the whole plane = Ap = 4*L^2
If the third point P3 falls on or within the circle, then the angles are all less than or equal to 90 degrees maximum (since if P3 is on the circle, the angle at P3 is always 90, but if P3 is just inside the circle, the angle at P3 exceeds 90, whereas if the third point is just outside the circle, the angle at P3 is less than 90.
If P3 falls within the half plane opposite to the half plane containing x, the triangle will certainly be OBTUSE If P3 falls within the rest of the half plane beyond x, the triangle will certainly be OBTUSE If P3 falls within the circle diameter x, the triangle will certainly be OBTUSE
Thus the condition that the points 0, x and P3 form an OBTUSE triangle is the SUM of the probabilities that (1) P3 falls within the halfplane opposite to the halfplane containing x, probability=H/Ap (2) P3 falls within the area of the rest of the half plane beyond x, probability (HS)/Ap (3) P3 falls within the area of the circle within the strip S, probability (Acirc/S)*(S/Ap)
Since the total probability must also depend on the probability where x occurs, P(x), we have
P(OBTUSE)= [H/Ap+(HS)/Ap+Acirc/H]*P(x)
which after substituting for S, H, AP and Acirc, simplifies to
P(OBTUSE)=[ 1  abs(x)*2*L/(4*L^2) +(pi*abs(x)^2/4)/(4*L^2)]*P(x)
The final answer depends on integrating over x, given that P(x)=dx/(2*L) for a unifirmly distributed x between L<x<L and noting that the integral is taken for L<x<L, which can be replaced by twice the integral taken over 0<x<L, since abs(x) osccurs in the integrand.
The result is
P(OBTUSE)=2/(2*L)integral([ 1  abs(x)*2*L/(4*L^2) +(pi*abs(x)^2/4)/(4*L^2)]dx, between x=0,L
which is
P(obtuse)=2/(2*L)*[LL^2*2*L/(2*4*L^2) + pi*L^3/(3*4)/(4*L^2)]
Which reduces to
P(OBTUSE)=[1(1/4)+pi/48]
So my final answer is 0.75  pi/48, and not 0.75.
 Tony Richards 'I think, therefore I am confused' Rutherford Appleton Lab ' UK '

