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Topic: Random Triangle Problem
Replies: 57   Last Post: Aug 17, 1997 10:51 PM

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 Robert Hill Posts: 529 Registered: 12/8/04
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 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)

Date Subject Author
7/16/97 Mike Housky
7/21/97 Bill Taylor
7/22/97 tony richards
7/24/97 Brian M. Scott
7/23/97 tony richards
7/23/97 T. Sheridan
7/24/97 Bill Taylor
7/24/97 Bill Taylor
7/25/97 Ilias Kastanas
7/23/97 Robert Hill
7/23/97 tony richards
7/27/97 Bill Taylor
7/24/97 Robert Hill
7/28/97 tony richards
7/30/97 Bill Taylor
7/30/97 tony richards
8/1/97 Bill Taylor
7/24/97 Robert Hill
7/24/97 Robert Hill
7/24/97 Robert Hill
7/25/97 Robert Hill
7/30/97 Bill Taylor
8/1/97 Charles H. Giffen
8/1/97 John Rickard
8/1/97 Chris Thompson
8/1/97 John Rickard
8/4/97 Bill Taylor
8/5/97 John Rickard
7/25/97 Charles H. Giffen
7/25/97 Charles H. Giffen
7/28/97 Hauke Reddmann
7/28/97 Robert Hill
7/28/97 Robert Hill
7/28/97 Robert Hill
7/29/97 tony richards
7/30/97 Keith Ramsay
7/30/97 tony richards
8/2/97 Keith Ramsay
7/29/97 tony richards
8/4/97 Bill Taylor
8/5/97 Charles H. Giffen
8/6/97 Terry Moore
8/7/97 Terry Moore
8/16/97 Kevin Brown
8/17/97 Kevin Brown
7/30/97 Robert Hill
7/31/97 tony richards
8/6/97 Terry Moore
7/31/97 John Rickard
7/30/97 Robert Hill
7/31/97 Robert Hill
7/31/97 Robert Hill
8/1/97 R J Morris
8/4/97 Robert Hill
8/4/97 Robert Hill
8/5/97 Charles H. Giffen
8/6/97 Robert Hill