Probability of |BC| < T Given Distances |AB| and |AC|

Asked Sep 06 '23 at 02:47

Active Sep 07 '23 at 23:07

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There are 3 unknown points in Euclidian space namely A,B and C. given distances |BC| and |AC|, what is the probability that |AB| < T.

For the 2d space using intersection of the circles and comparing the arc to circumference, I arrived at $\frac{\theta}{2\pi}$ where

$\theta = 2\arccos{\frac{|AC|^2+|BC|^2-T^2}{2|AC||BC|}}{2\pi}$

And for 3d using spheres somehow I got to: $\frac{\sin\left(\frac{\theta}{2}\right)}{2}$

So I'm not sure if I'm doing any of this correctly or the answer really is different for higher dimensions. I'm trying to find the general answer for n-dimensional space.

asked Sep 06 '23 at 02:47

optimum

What distribution are the points chosen from? – Command Master Sep 06 '23 at 05:00
@CommandMaster we can assume uniform distribution. – optimum Sep 06 '23 at 10:20
There's no uniform distribution on the entire euclidean plane – Command Master Sep 06 '23 at 10:25
@CommandMaster right. perhaps we can use uniform distribution with large enough bounds. (sorry, I'm not a mathematician, I don't even know how to formulate this problem correctly) – optimum Sep 06 '23 at 10:40

Probability of |BC| < T Given Distances |AB| and |AC|

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