Hmm, my asymptotic approximation is significantly less than yours; I get r(n) approaches sqrt(2 sqrt(3)/pi) * sqrt(n) as n approaches infty Unless my math skills have failed miserably.

On 10/24/08, Dan Asimov <dasimov@earthlink.net> wrote:

Kevin Buzzard posted this to sci.math.research, and I am taking the liberty of reposting it here:

<< The following came up in an undergraduate problem-solving group: given 6 distinct points in the plane we can consider the 6-choose-2 distances between pairs of distinct points. If M is the largest of these distances, and m is the smallest, then show that M/m>=sqrt(3).

In fact sqrt(3) isn't optimal, but I don't know what the optimal number is. One might ask what the answer is with n>=3 points in the plane: a compactness argument shows that there will be some r(n)>=1 such that M/m is always at least r(n) and furthermore such that r(n) is attained by some configuration of n points. . . .

6_C_2 = 15, which divides by 3 but not by 2. Assuming maximal symmetry in the extremum --- not always justified! --- suggests a convex hexagon with 6 sides of equal length 1, and 9 diagonals of equal length x. The Cayley-Menger determinant for a (flat) kite with sides 1,1,x,x and diagonals x,x, reduces to equation x^4 - 4x^2 +1 = 0, with solution (1 + sqrt3)/sqrt2 = 1.931852... [Intriguingly, another solution is the reciprocal of the above --- to what configuration might this correspond?] Any better offers for min(M/m) ? Fred Lunnon