
Re: Random Triangle Problem
Posted:
Jul 24, 1997 8:54 AM


In article <5r5375$p8u@newton.cc.rl.ac.uk>, tony richards <tony.richards@rl.ac.uk> writes: > eclrh@sun.leeds.ac.uk (Robert Hill) wrote: > >In article <5r2l8b$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).
I'm not happy about your wording. A set can't be an infinite plane and a bounded square at the same time. More seriously, there is an assumption here that the answer for a square is the same as "the answer" (whatever that means) for the whole plane. I have an informal argument (which I suspect I could make rigorous) that the probability of obtuseness for triangles whose vertices are uniformly distributed on any convex set of finite area is < 3/4, but I still suspect that the best answer for "a uniform distribution on the whole plane" is = 3/4. But that's like a statement about unicorns.
There is a second assumption in your argument that it's OK to take the first point as centre of the square, and a third that it's OK to orient the square so that a pair of sides is parallel to the line joining the first two points. I have evidence against the conjunction of these assumptions (see below).
I wonder if anyone has worked this out for points with coordinates that are independently distributed normal N(0,1)?
However, to continue ...
> 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
I think here you mean either "outside" rather than "within", or else you mean "greater" rather than "less", but your next words show that that's only a passing typo.
> (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)
Agreed, for each value of x (subject to previous dubious assumptions). > 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)
I don't see why the last term in the square brackets has become Acirc/H rather than Acirc/Ap. But again, this slight slip seems not to be propagated.
> 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.
Considering separately the three terms for an obtuse angle at the first, second and third points, I get that (conditionally on x) the probabilities are
obtuse at first point: 1/2 obtuse at second point: 1/2  x/(2L) obtuse at third point: pi x^2 / (16L^2) all angles acute: x/(2L)  pi x^2
which after integration over x become respectively
obtuse at first point: 1/2 obtuse at second point: 1/4 obtuse at third point: pi/48 all angles acute: 1/4  pi/48
Surely a correct answer would give equal probabilities for obtuseness at the three vertices.
 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)

