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Subject: "Another solution to geoprobability problem"     Previous Topic | Next Topic
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Prasanna Meda
Jul-23-06, 06:39 AM (EST)
"Another solution to geoprobability problem"
   The solution uses the bary centric co-ordinates. I think it is easy to map the possible areas and find the probability.

The total possible points that satisy the conditions are 0<=x<<1, 0<=y<=1, 0<=z<=1 and x + y + z = 1 is a traingle on the plane x+y+z =1 with x-y plane, y-z plane and x-z plane as sides. Its three vertices are (1,0,0),(0,0,1) and (0,1,0). Its side is sqrt(2) units. Its area is sqrt(3)/4*(sqrt(2)*sqrt(2))

The total possible points that satisfy the conditions 0<=x<=1/2, 0<=y<=1/2, 0<=z<=1/2, x + y + z = 1 is another trinagle with sides (1/2,1/2,0), (0,1/2,1/2) and (1/2,0,1/2). Its side is sqrt(2)/2 units and its area is sqrt(3)/4*sqrt(2)/2*sqrt(2)/2. This trinagle is actually formed with mid points of the sides of the above traingle.

The probability is thus 1/4, the ratio of areas of the traingles.

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Charter Member
1865 posts
Jul-24-06, 03:40 PM (EST)
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1. "RE: Another solution to geoprobability problem"
In response to message #0
   I would not call this "another solution". The difference between what you are saying and the page you refer to is two-fold:

  1. You observe the fact that the barycentric coordinates are nothing but the regular Cartesian coordinates in the plane x + y + z = 1. This is a very good observation.

  2. You explicitly compute the areas of the two triangles involved. But there is no need for that. As a medial triangle, the small one has 1/4 of the area of the big triangle.

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