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Prasanna Meda guest
Jul-23-06, 06:39 AM (EST)

"Another solution to geoprobability problem"

 The solution https://www.cut-the-knot.org/triangle/geoprobability.shtml 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.

alexb Charter Member
1865 posts
Jul-24-06, 03:40 PM (EST)    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: 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. 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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