Let $A,B,C$ be three points, chosen independently and uniformly at random on the circumference of a circle. Let $b(x)$ be the probability that at least one of the angles of the triangle $ABC$ exceeds $x\pi$. Show that $$ b(x)= 1-(3x-1)^2\ \ \ \ \ \mbox{if}\ \ \frac{1}{3}\leq x\leq \frac{1}{2};\\ b(x)= 3(1-x)^2 \ \ \ \ \ \mbox{if}\ \frac{1}{2}\leq x\leq 1. $$

My computation is as follows: Without loss of generality, we assume the circle is unit circle.
Let $A=(1,0),\ B=(1,\theta), $ and $C=(1,\phi)$ where $0<\theta<\phi<2\pi$. It is obvious that the joint density function $f(\theta,\phi)=(2\pi^2)^{-1}$ with $0<\theta<\phi<2\pi.$ Moreover, we have $$ \angle A=\frac{1}{2}(\phi-\theta),\ \ \angle B=\pi-\frac{\phi}{2},\ \ \mbox{and}\ \ \angle C=\frac{1}{2}\theta. $$ First, consider $x\in [1/3,1/2]$. I tried to compute the probability that all angles are less than $x\pi$ as follows: $$ \frac{1}{2}\theta \leq x\pi,\ \ \frac{1}{2}(\phi-\theta)\leq x\pi,\ \ \mbox{and} \ \ \pi-\frac{1}{2}\phi\leq x\pi. $$ Thus, we have $$ \theta \in [0,2x\pi],\ \ \mbox{and}\ \ \phi\in [2(1-x)\pi,2x\pi+\theta]. $$ Hence, $$ \int_{0}^{2x\pi}\int_{2(1-x)\pi}^{2x\pi+\theta}\frac{1}{2\pi^2}d\phi d\theta =\frac{1}{2\pi^2}\int_{0}^{2x\pi} [2x\pi+\theta-2\pi+2x\pi]d\theta = \\ \frac{1}{2\pi^2}\int_{0}^{2x\pi} [2\pi(2x-1)+\theta]d\theta = \frac{1}{2\pi^2}[(2\pi)(2x-1)(2x\pi)+2x^2\pi^2] =5x^2-2x. $$ Thus, I obtain $b(x)=1-(5x^2-2x)=-5x^2+2x+1.$ The result is much different to the the goal.

For the case $x\in [1/2,1].$ I analyzed as follows:

It is obvious that $\angle C=1/2\theta $ is impossible exceed $x\pi$, so we only need to consider angle $B$ and angle $A$. $$ \angle B>x\pi \Rightarrow \pi-\frac{1}{2}\phi>x\pi \Rightarrow \phi <2\pi(1-x). $$

Thus, we have $\angle B>x\pi\ \ $ if and only if $\ \ \phi\in [0,2\pi(1-x)]$ and $\theta\in [0,\phi].$ $$ \angle A>x\pi \Rightarrow \frac{1}{2}(\phi-\theta) >x\pi \Rightarrow \phi>2x\pi+\theta. $$ On the other hand, $\angle C\leq x\pi$, and hence $\theta\leq 2x\pi.$

Thus, we have $\angle A>x\pi \ \ $ if and only if $\ \ \phi\in [2x\pi+\theta,2\pi]$ and $\theta\in [0,2x\pi].$ For this case (x\in [1/2,1]), I also got the the different $b(x)$.

I wonder is there any part I did wrong? Or, is there any good suggestion how to work it? Thanks.


It often helps with this kind of problem to draw a diagram. For example, here is a diagram for the first case:


Here, $\theta$ is the horizontal axis, $\phi$ is the vertical axis, the region of all $(\theta, \phi)$ with $0 \le \theta \le \phi \le 2\pi$ is in blue, and the smaller region in red is the region satisfying $$\frac{1}{2}\theta \leq x\pi,\ \ \frac{1}{2}(\phi-\theta)\leq x\pi,\ \ \mbox{and} \ \ \pi-\frac{1}{2}\phi\leq x\pi.$$ (Each of the individual inequalities is shown by a dashed line.)

I think this makes clear at least one of your mistakes: your integral over $\theta$ is starting at $\theta=0$, but the red region doesn't actually reach $\theta=0$, so if everything else is correct you'll end up with the difference between the red triangle, and the small blue triangle to its left formed by two dashed lines.

But with the diagram, we don't need to do an integral at all. The red region is an isosceles triangle with vertices $((2-2x)\pi, (2-2x)\pi)$, $(2x\pi, (2-2x)\pi)$, and $(2x\pi, 4x\pi)$. Therefore one of the shorter sides has length $(2-6x)\pi$, and its area is $\frac12 \big((2-6x)\pi\big)^2 = 2\pi^2 \cdot (1-3x)^2$.

Divide that by $2\pi^2$, the area of the blue triangle, and we conclude that the probability that no angle exceeds $2\pi$ in this case is $(1-3x)^2$, and $b(x) = 1 - (1-3x)^2$ is the probability we want.

For the second case, the diagram is:


Here, it's easier to directly count the blue area, instead of counting the red area and subtracting, as we did in the previous case. There are three triangles, which turn out to have area $(1-x)^2 \cdot 2\pi^2$ each, corresponding to the probability that each of the angles is bigger than $x\pi$.

  • $\begingroup$ Thanks! It is very clear by seeing the pic. $\endgroup$ – Pei-Lun Tseng Apr 17 '17 at 2:50

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