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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?

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  • $\begingroup$ Uniformly distributed along the perimeter, I assume? $\endgroup$ – Frpzzd Nov 13 '18 at 0:45
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    $\begingroup$ With aid of Mathematica, the average distance between two independent points chosen unformly at random from the perimeter of a unit square is $$\frac{1}{12} \left(3+\sqrt{2}+5 \operatorname{arcsinh}(1)\right) \approx 0.73509012478923418125\cdots $$ $\endgroup$ – Sangchul Lee Nov 13 '18 at 0:59
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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 $p_i.$

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