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Old 2007-04-14, 07:09   #1
Kees
 
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Default 4 Points in a Plane

Choose 4 points at random in a plane.
What is the probability that the convex hull of these 4 points is a triangle ?
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Old 2007-04-14, 08:23   #2
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Is the plane infinite?

What is the type of random distribution?
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Old 2007-04-14, 08:51   #3
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Is this the same as asking whether they can constitute
the vertices of a quadrilateral whose interior angles are all
< 180 degrees?
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Old 2007-04-14, 08:51   #4
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the plane is infinite, the distribution is that every point in the plane is as likely to be selected
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Old 2007-04-14, 08:56   #5
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Quote:
Is this the same as asking whether they can constitute
the vertices of a quadrilateral whose interior angles are all
< 180 degrees?
Not sure what you mean exactly, but 4 points in a plane have either a triangle or a quadrilateral as convex hull.
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Old 2007-04-14, 09:57   #6
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I say it's evens.
Take three points at random. Produce the three sides of the
triangle in both directions. This divides the plane into seven
regions: the triangle itself, 3 unbounded regions adjoining an edge
of the triangle, and 3 unbounded regions with one vertex. The two types of infinite region each span 180 degrees.
The fourth point has equal probability of lying in each type of infinite
region.
(The hull is a triangle iff the fourth point lies in a "vertex"
type region or the triangle itself)
(I shudder to think what Silverstone will make of this )

David

Last fiddled with by davieddy on 2007-04-14 at 10:04
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Old 2007-04-14, 10:03   #7
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My solution is essentially the same and I too fear that it is not very rigourous
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Old 2007-04-14, 10:46   #8
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If we consider a sphere rather than a plane, does this
clarify or obfuscate the problem?

David
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Old 2007-04-14, 11:19   #9
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Quote:
Originally Posted by Kees View Post
My solution is essentially the same and I too fear that it is not very rigourous
I suspect that Retina may have been anticipating these problems.
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Old 2007-04-14, 14:57   #10
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Quote:
Originally Posted by Kees View Post
the plane is infinite, the distribution is that every point in the plane is as likely to be selected
In that case, I think the probability approaches zero (P --> 0).
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Old 2007-04-14, 21:33   #11
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Quote:
Originally Posted by retina View Post
In that case, I think the probability approaches zero (P --> 0).
In the context of my analysis, I think you are referring
to the probability of the fourth point lying inside the triangle
formed by the other three.
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