Geometry : Probability that Three Points on a Circle will form a Right-Triangle
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If n points are equally spaced on the circumference of a circle, what is the probability that three points chosen at random will form a right triangle?
I know that for us to have a right triangle, the two points should form the diameter of the circle. What I have done is that I divided the problem into two sections.
Section I: n even
Section II: n odd
When n is even it is easier to find the probability but I can't find the right answer (both for n= even and n= odd).
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Solution Summary
The Probability that Three Points on a Circle will form a Right-Triangle is investigated.
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Since the n points are equally spaced on the circumference of a circle, we can suppose the n points are p_1,p_2,...,p_n in clockwise order. From the condition, we know all the distance d(p_i,p(i+1)) (1<=i<=n) are the same. (When i=n, i+1=1)
We know, if we want to form a right triangle from three points, two of them should form a diameter of the circle.
Now we consider two cases:
Case 1: n is odd. Then I claim any three points can ...
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