# Probability of obtuse triangles formed on a circle

There are 16 equally spaced points on the circumference of a circle. If 3 points out of these 16 points are selected randomly, What is the probability that they will form an obtuse angled triangle?

• Fix a point. Because a circle with equally spaced points is perfectly symmetrical, it does not matter which one you start with. Assume that point is one of the three. Now, to choose the other two points, there are $\dbinom{15}{2} = 105$ different ways of choosing them. Determine how many pairs result in triangles with obtuse angles. The answer will be that number over 105. – InterstellarProbe Apr 17 at 17:16
• yeah, I was thinking the same but I am not able to count the number of obtuse triangles in these 105 triangles. – James Joy Apr 17 at 17:30
• Under what circumstances will the angle be obtuse? Determine what circumstances will make the angle obtuse and look for patterns. For instance, if all three points are right next to each other, it is obtuse. If they are all spread out, every angle is acute. Move one point until it goes from acute to obtuse (if ever). – InterstellarProbe Apr 17 at 17:33
• If it may help to check your result: by brute force by program (using trigonometry for the angles): the probability to get an obtuse triangle is $3/5$ (to be clear, by obtuse I mean one angle is $>\pi/2$). – Jean-Claude Arbaut Apr 17 at 17:34
• If you pick $n$ equally spaced points and let $n\to\infty$, then the probability converges to $3/4$. Found by numerical experiment, but see also Richard K. Guy, There Are Three Times as Many Obtuse-Angled Triangles as There Are Acute-Angled Ones. Moreover, for $n=4k$ points, the probability seems to be $\dfrac{3}{4}\cdot\dfrac{n-4}{n-1}$. All of this should be proved carefully though. – Jean-Claude Arbaut Apr 17 at 20:25