As a warmup however, I did calculate the average distance between points on the boundary of an equilateral triangle with unit side lengths (measuring distances along the boundary, as if the sides of the triangle were roads).

The average point-to-point distance is ¾, which is less than the length of a side. For details, see this two-pager.

Waking up fresh this morning, I think the mean distance is easy to calculate as follows. Once the first point is chosen, the second point is equally likely to be closer when measured clockwise or closer when measured counter-clockwise. If closer measured clockwise, all distances up to 3/2 are equally likely; if closer measured counter-clockwise, all distances up to 3/2 are equally likely. Therefore once the first point is chosen, all distances up to 3/2 are equally likely, and the mean distance is

\[\int_0^{\frac{3}{2}}{v\cdot\left(\frac{2}{3}\,dv\right)} = \frac{3}{4}\,.\]

This reasoning generalizes to any triangle if we replace 3/2 by the semiperimeter

\[{\rm mean\ distance} = \int_0^s{v\cdot\left(\frac{1}{s}\,dv\right)} = \frac{1}{2}s = \frac{1}{4}p\]

***

Waking up fresh this morning, I think the mean distance is easy to calculate as follows. Once the first point is chosen, the second point is equally likely to be closer when measured clockwise or closer when measured counter-clockwise. If closer measured clockwise, all distances up to 3/2 are equally likely; if closer measured counter-clockwise, all distances up to 3/2 are equally likely. Therefore once the first point is chosen, all distances up to 3/2 are equally likely, and the mean distance is

\[\int_0^{\frac{3}{2}}{v\cdot\left(\frac{2}{3}\,dv\right)} = \frac{3}{4}\,.\]

This reasoning generalizes to any triangle if we replace 3/2 by the semiperimeter

*s*:\[{\rm mean\ distance} = \int_0^s{v\cdot\left(\frac{1}{s}\,dv\right)} = \frac{1}{2}s = \frac{1}{4}p\]

where

*p*is the perimeter of the triangle.
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