# arbitrary three points on a circle

Three points $A,B,C$ are chosen randomly on a circle.Find the probability that $\angle ABC$ is less than $\theta \in (0,\pi)$. My method of solving this problem is this:

Fix a point $B$ on the circle and then define $a=2\pi s$ and $b=2\pi t$ , $s,t\in [0,1]$.Then $\angle ABC \le \theta$ if and only if $|s-t| \le \frac{\theta}{\pi}$. Define a set $K=\{(s,t)\in [0,1]^2:|s-t| \le \frac{\theta}{\pi}\}$ in $[0,1]^2$.

Is it true that $P(\angle ABC \le \theta)=\int \int_K dsdt$?

Can we solve this problem without integration?

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I think I saw an elementary solution to this problem in Yaglom and Yaglom, "Challenging Mathematical Problems with Elementary Solutions", not sure since it was many years ago when i last read it.... –  N. S. Jun 21 '12 at 16:04
See also this related question: math.stackexchange.com/questions/115895/… –  yasmar Jun 22 '12 at 7:08

Assume without loss of generality that the affixes of $A$, $B$ and $C$ are $\mathrm e^{\mathrm iu}$, $1$ and $\mathrm e^{\mathrm iv}$ with $u$ and $v$ in $(-\pi,\pi)$, and let $O$ denote the center of the circle. The triangles $BOA$ and $BOC$ are isocele hence the angles $\angle OBA$ and $\angle OBC$ are $\frac12\pi-\frac12u$ and $\frac12\pi-\frac12v$. In particular the angle $\angle ABC$ is $\frac12(u-v)$.

The nonoriented angle at $B$ is $\frac12|u-v|$ in $(0,\pi)$ with $u$ and $v$ i.i.d. uniform in $(-\pi,\pi)$ hence its distribution has density $\frac2{\pi^2}(\pi-\theta)$ on $0\leqslant \theta\leqslant\pi$. In other words, the probability that this angle is at least $(1-a)\pi$ is $a^2$, for every $0\leqslant a\leqslant 1$.

Edit: A simple approach to the CDF is to note that, after dividing everything by $\pi$, one looks for the distribution of $t=|x-y|$ where the random point $(x,y)$ is uniform in the square $[0,1]^2$.

For every $0\leqslant r\leqslant 1$, the event $[t\geqslant r]$ corresponds to $(x,y)$ falling into the triangle with vertices $(r,0)$, $(1,0)$ and $(1,1-r)$, or into the triangle with vertices $(0,r)$, $(1-r,1)$ and $(0,1)$. Each triangle is half of a square with side $1-r$ hence the sum of their areas is $(1-r)^2$. The PDF of the angle normalized by $\pi$ is minus the derivative of this, that is, $2(1-r)$.

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In fact, I am not quite familiar with probability.I do not quite understand how you can show that the distribution has density $\frac{2}{\pi^2}(\pi−x)$.Can you further explain this? Thank you. :) –  Ben Jun 21 '12 at 17:16
See Edit.   –  Did Jun 21 '12 at 17:28

(This answer was wrong. I have deleted it.)

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However,B is not always on the major arc of AC,so I am not sure whether angle at B=angle AOC over 2. –  Ben Jun 21 '12 at 23:46
@BenLi: See my edit. –  Christian Blatter Jun 22 '12 at 8:02
To assume that C, B and A are arranged counterclockwise seems to modify the distributions of A and C. The original uniform distribution of A should be biased towards the small values of the angle BOA, and the other way round for C. –  Did Jun 22 '12 at 8:18

Browlowski can.

Its less then pi iff a,b,c lie in the same halfcircle. Pick A, then pick C, then by symmetry you have a 50% chance B falls in the same halfcircle, and by symmetry 50% chance it falls on the right side of C. So 50%*50%=25%.

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The answer has to depend on $\theta$. You're assuming $\theta = \pi$. –  dfan Jun 21 '12 at 17:18
ah, right you are –  user1708 Jun 21 '12 at 17:19