Spreading points evenly in plane

[shybishie]

Place 6 points in the plane, such that ratio of maximum distance / minimum distance (over these points) is as small as possible. The question is - what is the smallest ratio possible, and can we prove this is a tight bound? The following is my attempt for 6 (and fewer points):

1) For three points, we can clearly achieve max dist / min dist = 1 (equilateral triangle)

2) For four points, I tried 2 configurations - a point inside a triangle (couldn't do better than \sqrt{3} doing that though, using an equilateral triangle and it's centroid) . Also, I got a better result with a square (I got \sqrt{2} , with the maximum length side being the diagonal). I think that should be improvable though?

3) For five points, the best I got was with a regular pentagon (about 1.62).

4) For six points, a regular hexagon was NOT the best distribution - that gave a ratio of 2 between largest and smallest distance. The best I got was a regular pentagon, with the sixth point at it's intersection of angle bisectors - about a ratio of 1.9. Any way to improve this, and (more trickily) prove it's tight? My source says \sqrt{3} is the tight bound?

I guess where I'm most stuck is proving a lower bound; i.e, you can't do better than x - I'm not sure how to get started there, even for the n = 4 case. If anyone has ideas or strategies, I'd really appreciate.

P.S: This isn't a homework or coursework problem - just a random puzzle from Rutgers problem solving seminar website.

 

[Eynstone]

We can use Appolonious circles. Given two points A,B ,moving a third point P such that
PA/PB is invariant & P is as 'evenly' spaced with respect to other points as possible.
It follows that the configuration with the minimal no. of different distances is the optimal.
( We could keep equalizing the distances reducing the max/min ratio).

 

[shybishie]

Thank you Eynstone, that was very helpful :).

 

[Tinyboss]

What Eynstone said is better, but I'll go ahead and post the idea I had, too. We can place the first two points arbitrarily. Then assigning the value of the max/min ratio to each possible configuration of the remaining n-2 points defines a continuous function from R^(2n-4) to R (almost--you have to either adjoin infinity to the range or exclude configurations with coincident points). Then you might be able to use calculus tools to identify critical points.

 

[D H]

That function will be continuous but it will not be differentiable everywhere. There will be cusps. Even worse, the optimal configuration will be precisely at a cusp.