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

 Messages: [ Previous | Next ]
 tony richards Posts: 164 Registered: 12/8/04
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 '

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