An Expected Value Problem

Hi All... Recently I thought of a problem and tried to solve it. To my surprise, I ended up with a closed form solution for the same which I would like to share with you by this post.

Consider a rectangle $ABCD$ with the longest side $AB=CD=a$ units and shorter side $BC=DC=b$ units. Now pick a random point in this rectangle and a draw a straight line from this point at a random angle until the line meets the edge of the rectangle. What will be the expected value of this line?

We can attack the problem head-on with multiple integrals. Let $L$ be the length of the line. Then we have,

$\mathbb{E}(L)=\displaystyle\frac{1}{ab}\frac{1}{2\pi}\int\limits_{0}^{a}\int\limits_{0}^{b}\int\frac{y}{\sin t}\,dt\,dy\,dx$

The limits of the innermost integral have been left out purposely. We have to decompose the innermost integral. Let's do it this way. Consider the point $(x,y)$. We'll call it $P$.

Now draw a perpendicular from this point to each of the four sides of the rectangle. Let the perpendicular meet the side $X$ at $P_X$. Also join this point to each of the four vertices of the rectangle. This splits the entire rectangle into eight regions. 

Consider the integral $I(a,b)$ (ignoring constant factors for time being) for the 'random lines' that end up in region $PAP_{AB}$

$I(a, b)=\displaystyle\int\limits_{0}^{a}\int\limits_{0}^{b}\int\limits_{\tan^{-1}(y/x)}^{\pi/2}\frac{y}{\sin t}\,dt\,dy\,dx$

But reflecting the 'random lines' about the $y$-axis, this integral also represents the $PP_{AB}B$, reflecting about the $x$-axis, this integral represents region $PP_{CD}D$, reflecting w.r.t both the axes it represents region $PCP_{CD}$. Solving this one integral covers four of the eight regions.

This is integral is pretty simple to solve with standard tables (or atmost with Mathematica). We get,

$I(a,b)=\displaystyle\frac{1}{6}\left(2b^3-a^3+d^3-3b^2d+3ab^2\ln{\frac{a+d}{b}}\right)$

where $d$ is the length of the diagonal of the rectangle.

Now for the other regions. We don't have to solve anything separately. Just interchange the values of $a$ and $b$. This amounts to rotating the rectangle by $90$ degrees and reasoning as before for the four other regions.

Note the nice thing that in going over the eight regions, we have made the full $2\pi$ radians possible for the 'random line'. So finally we have,

$\mathbb{E}(L)=\displaystyle\frac{4I(a,b)+4I(b,a)}{2\pi ab}$

Simplifying things we finally we end up with,

$\mathbb{E}(L)=\displaystyle\frac{a^3+b^3-d^3}{3\pi ab}+\frac{a}{\pi}\ln{\frac{b+d}{a}} + \frac{b}{\pi}\ln{\frac{a+d}{b}}$

(or)

$\mathbb{E}(L)=\displaystyle\frac{a^3+b^3-d^3}{3\pi ab} +\frac{a}{\pi}\text{csch}^{-1}\left(\frac{a}{b}\right)+ \frac{b}{\pi}\text{csch}^{-1}\left(\frac{b}{a}\right)$

The rectangle had a lot of symmetry that we were able to exploit. I'm trying to do the same for a given arbitrary triangle but it seems so very difficult with complicated integrals cropping at all places. I'll update if end up with something.

Until then,

Yours aye

Me