TITLE: The minimum-perimeter triangle of three sets of points
QUESTION [8 upvotes]: If $X$ and $Y$ are two sets of $n$ independent, uniformly sampled points in the unit square, then standard methods can show that the expected minimum distance between points in $X$ and $Y$ is proportional to $1/n$, that is, $$E(\min_{i,j}\|x_i - y_j \|)\sim 1/n$$as $n\to\infty$.  Is there anything similar that can be said when we have three sets of points $X,Y,Z$, and we look for the triangle whose perimeter is shortest?  I.e., $$E(\min_{i,j,k}\|x_i-y_j\|+\|y_j-z_k\| + \|z_k - x_i\|)$$?

REPLY [9 votes]: Partition the unit square into small squares of area roughly $a$. Your question is equivalent to asking for which $a$ do we typically see about 1 small square with points from each of $X$,$Y$ and $Z$? this probability is roughly $(an)^3$ since the probability of seeing a point from each set is about $an$ and they're independent. There are $1/a$ small squares so the expected number of squares with points from each set is roughly $(an)^3/a=a^2n^3$. So if we want this to be roughly 1 we need to set $a=n^{-3/2}$. The diameter of a small square is therefore $n^{-3/4}$.
Generalizing for $k$ different sets and dimension $d$ yields $n^{-\frac{k}{d(k-1)}}$.