A probability problem by Tai Ching Kan

This was a really enjoyable problem!

For convenience, I scaled the entire problem up $\frac{4}{3}$ times, so that the smaller square has side $1$ . The final answer, of course, doesn't change.

Let the two points, $A$ and $B$ , have coordinates $a_1,a_2$ and $a_3,a_4$ respectively. Note that $a_1,a_2,a_3,a_4$ are uniformly random variables between $0$ and $1$ .

A natural first step is to evaluate the coordinates of the remaining two points, $C$ and $D$ .

The coordinates of the center of the square are $\displaystyle\frac{a_1+a_3}{2},\frac{a_2+a_4}{2}$ . To find, say, the $x$ coordinate of $C$ , one must add $\displaystyle\frac{\sqrt{(a_1-a_3)^2+(a_2-a_4)^2}}{2}\cdot cos{\theta}$ = $\displaystyle\frac{\sqrt{(a_1-a_3)^2+(a_2-a_4)^2}}{2}\cdot\frac{(a_4-a_2)}{\sqrt{(a_1-a_3)^2+(a_2-a_4)^2}}=\displaystyle\frac{a_4-a_2}{2}$ to the center point's $x$ coordinate, $\displaystyle\frac{a_1+a_3}{2}$ , as seen in the figure.

Finding the other three coordinates we need in a similar fashion, we find they are:

$\displaystyle\frac{a_1+a_2+a_3-a_4}{2}$ , $\displaystyle\frac{a_1+a_2+a_4-a_3}{2}$ , $\displaystyle\frac{a_1+a_3+a_4-a_2}{2}$ and $\displaystyle\frac{a_2+a_3+a_4-a_1}{2}$ . In order for the formed square to be entirely within the one defined by $0<x,y<\frac{4}{3}$ , these four values have to all be inside $(0,\frac{4}{3})$ .

We arrive at the following system:

$\begin{cases}0<a_1+a_2+a_3-a_4<\frac{8}{3}\\0<a_1+a_2+a_4-a_3<\frac{8}{3}\\0<a_1+a_3+a_4-a_2<\frac{8}{3}\\0<a_2+a_3+a_4-a_1<\frac{8}{3}\end{cases}$ .

Note that only one of these four constraints can be broken at a time, as you can show easily (the difference between any two of the values is at most $2$ and if we assume any two can be less than $0$ or more than $\frac{8}{3}$ , adding the two equations immediately leads to a contradiction). This will be important later.

Finding the probability density function for $a_1+a_2+a_3-a_4$ is rather straightforward using $\href{https://en.wikipedia.org/wiki/Convolution_of_probability_distributions}{convolution}$ of PDF's:

$\displaystyle P(a_1+a_2+a_3-a_4=x)=\begin{cases}0\text{ if }x<-1\\\frac{(x+1)^3}{6}\text{ if }-1<x<0\\\frac{-3x^3+3x^2+3x+1}{6}\text{ if }0<x<1\\\frac{3x^3-15x^2+21x-5}{6}\text{ if }1<x<2\\\frac{(3-x)^3}{6}\text{ if }2<x<3\\0\text{ if }3<x\end{cases}$

So the probability a constraint is satisfied is:

$\displaystyle\int_{0}^{\frac{8}{3}} P(x)=\frac{931}{972}$

Thus the probability a constraint is not satisfied is $1-\frac{931}{972}=\frac{41}{972}$ .

Since only one of the constraints can be broken at a time, the probability any of them is broken is simply $4\cdot\frac{41}{972}=\frac{164}{972}$ , making our answer $1-\frac{164}{972}=\frac{808}{972}=\frac{202}{243}$ .