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.
 A: I'm assuming that the question means that three points are randomly picked on a circle, distributed uniformly by angles. In this case, having a 120 degree angle or greater requires the "central angle" to be 240 degrees or greater. Essentially, two-thirds of the circle have to be in one "piece" of the three-piece cut.
This is equivalent to the following problem: Cut the segment [0,1] at two random points. What is the probability that the largest segment now has length at least 2/3?
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 $$P={1\over 3}.$$
