My intuitions say that the maximal sum of radii for circles within a circle will approach, not sqrt(n), but sqrt(An), where A = pi sqrt(3)/6 ~ .9069 is the ration of the area inside the circles to the total area in a hexagonal close-packing. To get a radius-sum of sqrt(n), you'd want n circles of radius 1/sqrt(n). But those will have total area pi, the same as the area of the bounding circle, while my intuition says that when you have lots of small circles, they will be roughly the same size in roughly a hexagonal close-packing, so they won't have total area more than A pi. So if there are n circles, each should have area A pi/n, or radius sqrt(A)/sqrt(n), and n of them have total radius sqrt(nA). Andy Latto andy.latto@pobox.com On Fri, Feb 19, 2016 at 3:07 PM, Charles Greathouse <charles.greathouse@case.edu> wrote:
Maybe it's obvious, but I'd just like to know if the maximal sum of the radii of n circles in a circle of unit radius approaches its upper bound sqrt(n). (Or have I been caught in the sort of trap that inspired this conversation?)
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Fri, Feb 19, 2016 at 9:08 AM, Veit Elser <ve10@cornell.edu> wrote:
On Feb 19, 2016, at 8:48 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I'm a tad suspicious: are the circle packings on Friedman's page proved to be maximal, or are they merely the best known so far? There is no link to any proofs, or key to diameters … WFL
I have a conjecture that the maximum-sum-of-diameters packings have triangulated contact graphs. If that’s true, one could enumerate the graphs and calculate the packing for each one to see which is maximal.
-Veit
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