Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
It's interesting but somehow laborious !!!. –  Felix Marin Mar 23 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 at 23:45

1 Answer 1

up vote 1 down vote accepted

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 $$P={1\over 3}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.