# Probability that a triangle has an angle greater than 120 degrees

We've got a circle and we draw $3$ points, which form a triangle. Question: what is the probability that its greatest angle has more than $120$ degrees? Well, I have no idea how to do it. I know some methods of proving the probability that this triangle is obtuse, but it's a different problem... I thought I may think about central angles, but it doesn't work... I would appreciate your help.

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It's interesting but somehow laborious !!!. – Felix Marin Mar 23 '14 at 22:49
The probability that you are looking for does not depend on the placement of the first two point? Anyway if you suppose that it does not depend on this, the probability will be with direct relation to the set of point that form a triangle with the first two points and has a angle that are greater than 120. – npisinp Mar 23 '14 at 23:45

I'm not sure, but I think this gives the answer: Assume the first cut is somewhere in [0,1/2], say at the point $t$. If $0\le t\le 1/3$ then it is possible to achieve the condition. For fixed $t$, we would need the second cut to be somewhere in $[0,1/3]$ or $[t+2/3,1]$. So the probability, as a function of t, is $$P(t)=(1/3)+(1-(t+2/3))=2/3-t$$ To get the "average" probability, we integrate over valid choicse of t and divide by the length of the interval of integration: $$3\int_0^{\frac{1}{3}}{2/3-t}dt=3 \left(\frac{2}{9}-\frac{1}{18} \right)={1 \over 2}$$
EDIT: To get the final answer, we have to multiply by 2/3, which is the change that t actually lands in the appropriate range. Therefore the probably is $$P={1\over 3}.$$