Re: [math-fun] Distance ratio question
By the way, Kevin Buzzard & I just realized that this M/m problem is (trivially) equivalent to the "Penny Problem": -- Given n non-overlapping* unit disks in the plane, what is the smallest R = R(n) for which they can be arranged inside of a disk of radius R ? Neil, did you do any numerical work on this question? --Dan _____________________________________ * other than possible tangencies. _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
On 10/25/08, Dan Asimov <dasimov@earthlink.net> wrote:
By the way, Kevin Buzzard & I just realized that this M/m problem is (trivially) equivalent to the "Penny Problem": --
Given n non-overlapping* unit disks in the plane, what is the smallest R = R(n) for which they can be arranged inside of a disk of radius R ?
I think not. According to http://mathworld.wolfram.com/CirclePacking.html the optimal packing of the latter type arranges the 6 discs with centres on a regular hexagon, for which M/m = 2. For the configuration I descibed earlier, M/m = 1.931852 . WFL
Apologies --- I now realise that my configuration was not symmetrical as claimed! So M/m does have lower bound 2, etc etc. WFL On 10/25/08, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 10/25/08, Dan Asimov <dasimov@earthlink.net> wrote:
By the way, Kevin Buzzard & I just realized that this M/m problem is (trivially) equivalent to the "Penny Problem": --
Given n non-overlapping* unit disks in the plane, what is the smallest R = R(n) for which they can be arranged inside of a disk of radius R ?
I think not.
According to http://mathworld.wolfram.com/CirclePacking.html the optimal packing of the latter type arranges the 6 discs with centres on a regular hexagon, for which M/m = 2.
For the configuration I descibed earlier, M/m = 1.931852 .
WFL
participants (2)
-
Dan Asimov -
Fred lunnon