On Fri, Jan 27, 2017 at 3:58 PM, Julian Ziegler Hunts <julianj.zh@gmail.com> wrote:
A little. Take six disks connected in an arc, with constant angles between any consecutive three. Clearly, by varying that angle, we can make two disks fit across as they do in the picture. What angle makes this happen? The secant disks are symmetric with the two disks in the middle of the arc (leftmost disks in the picture), so those angles are the same. And the ends form equilateral triangles. So the total of the interior angles of the beehive is 6*θ - 2π/3=4π, and θ=7π/9, the interior angle of a nonagon.
Nice!
Of course, that doesn't really tell you why the angle happens to be the same as that of a nonagon. But it does prove by elementary geometry that they fit.
Julian
Re: [math-fun] They fit! Date: 2017-01-27 14:30 From: James Buddenhagen <jbuddenh@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Reply-To: math-fun <math-fun@mailman.xmission.com>
Similar things for n=11 and n=13: http://www.buddenbooks.com/jb/pack/circle/n11.htm
Perhaps this should be named Melissen's Theorem. I missed the swap @ 3 o'clock. And that solution looks to me quite unremarkable by comparison. Presumably, the possibile swap should be just as surprising as the secant pair fitting perfectly. But only the latter jumps out at me. Can the swap finish Julian's proof?
http://www.buddenbooks.com/jb/pack/circle/n13.htm I don't remember ever seeing two optimal solutions, one loose, one tight. Did you really mean i^2 = 1? --rwg
On Fri, Jan 27, 2017 at 4:16 PM, Bill Gosper <billgosper@gmail.com> wrote:
ccc11.gif <http://www2.stetson.edu/~efriedma/cirincir/ccc11.gif> seems to assert that two unit disks precisely fit within a circle of nine. After much bovoparturition, Mathematica agrees! Can anyone provide some insight? --rwg