Suprise! I screwed up again. Oh, well. Let's see if I can avoid screwing up the 2nd time and get something like what Gareth & Tom did: Of course, the right thing to do is look at at circle packed with a triangular array of points. WLOG say the circle is radius 1. Then M = max dist. ~ 2. As for m = min dist., let that be the side of each triangle in the array. Then each triangle's area is m^2 sqrt(3)/4, while the circle's is pi, so the number of triangles F ~ 4pi/(m^2 sqrt(3)). Here n is the number V of vertices, and each vertex (if the array is oriented so some edges are horizontal) can be made to correspond to 2 triangles -- the ones directly above & below it. Thus F ~ 2V = 2n. Thus 2n ~ 4pi / (m^2 sqrt(3)) and so 1/m ~ sqrt( n sqrt(3) / (2pi) ), so M/m ~ 2/m ~ sqrt(2 sqrt(3)/pi) * sqrt(n), which seems to agree with Tom's but is twice what Gareth wrote. So, Tom & I win, two to one. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele