In article <firstname.lastname@example.org>, tony richards <email@example.com> writes: > firstname.lastname@example.org (Robert Hill) wrote: > >In article <email@example.com>, tony richards <firstname.lastname@example.org> writes: > >> email@example.com (Bill Taylor) wrote: > >> >Mike Housky <firstname.lastname@example.org> 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 half-plane 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 = (L-abs(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 half-plane opposite to the half-plane containing x, > probability=H/Ap > (2) P3 falls within the area of the rest of the half plane beyond x, > probability (H-S)/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+(H-S)/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)*[L-L^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)