Cfrgtkky

Was just wondering, no special reason.

Someone asked the corresponding question about a square, so this question is a generalization of that question: What is the average distance of two points chosen uniformly on a unit square?

I will show you how to do it with integrals, but I'll leave it up to you to evaluate the integrals.

Suppose the rectangle has length $l$ and width $w$, and that points $X$ and $Y$ are chosen uniformly along its perimeter. We have a total of $5$ cases:

• Both points are on the same side of length $l$, with probability $p_1$
  


• Both points are on different sides of length $l$, with probability $p_2$
  


• Both points are on the same side of length $w$, with probability $p_3$
  


• Both points are on different sides of length $w$, with probability $p_4$
  


• One point is on a side of length $l$ and one is on a side of length $w$, with probability $p_5$
  

Consider the first case. If $x,y$ are the respective distances of $X$ and $Y$ from a given vertex of the side they are on, then their distance is $|x-y|$, and so the average distance is
$$I_1=frac{1}{l^2}iint_{[0,l]^2} |x-y|dxdy$$
Then consider the second case. Let $x,y$ be the respective distances of $X$ and $Y$ from a given side of length $w$ of the rectangle. Then the distance between $x$ and $y$ is given by $sqrt{(x-y)^2+w^2}$, and so the integral for their average distance is
$$I_2=frac{1}{l^2}iint_{[0,l]^2} sqrt{(x-y)^2+w^2}dxdy$$
Because the sides of length $l$ and those of length $w$ are indistinguishable in the context of the problem, the respective averages for case 3 and case 4 are analogously
$$I_3=frac{1}{w^2}iint_{[0,w]^2} |x-y|dxdy$$
$$I_4=frac{1}{w^2}iint_{[0,w]^2} sqrt{(x-y)^2+l^2}dxdy$$
Finally, for the fifth case, let $x$ and $y$ be the respective distances of $X$ and $Y$ from the vertex in common between their sides. Then, by the pythagorean theorem, the distance between them is $sqrt{x^2+y^2}$, so the integral is
$$I_5=frac{1}{lw}int_0^l int_0^w sqrt{x^2+y^2}dxdy$$
Now, the expected distance is simply
$$p_1 I_1+p_2I_2+p_3I_3+p_4I_4+p_5I_5$$
and so I'll leave it up to you to compute the integrals and the probabilities $p_i.$