Moving Dots 2

First, let $X_1$ denote the smaller angle of the two and find probability of $X$ to be smaller than $X_1$ . Make a sample space of $A$ and $B$ ( $0≤A≤2π, 0≤B≤2π$ ). To make $X$ smaller than $X_1$ , $|A-B|≤X_1$ or $|A-B|≥2π-X_1$ . Let it be specified it on the graph.

The mass of the dark region equals to $4π^2-(2π-X_1 )^2+(X_1)^2=4πX_1$ .

So the probability $P(X≤X_1)$ is $P(X≤X_1)=\frac{4πX_1}{4π^2}=\frac{X_1}{π}$ and the probability density function $f(x)=1/π$ .

$E(X)=\int_{0}^{π}\frac{x}{π} \ dx=\frac{π}{2}, E(X^2 )=\int_{0}^{π}\frac{x^2}{π} \ dx=\frac{π^2}{3}$ Therefore, $Var(X)=E(X^2 )-{{E(X)}}^2=\frac{π^2}{12}$ . Thus, $\Large \frac{π^2}{Var(X)}=\boxed{12}.$

The first point is chosen randomly, and its value doesnt really matter. The second point is chosen randomly, and due to symmetry, every angle between [0,pi] has exectly the same probability to occur. Thus said, we conclude that X~Uni[0,pi]. (In words, X distributrs uniformally on the values between 0 and pi). As we know, for Y~Uni[a,b] , we get that Var(Y)= ((a+b)^2 )/ 12. Thus , Var(X)=((0+pi)^2)/12 = (pi^2)/12. (Sorry for the lack of use of formulas, dont really now how to type formulas here)