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Re: how many coins?
Posted:
Apr 8, 1995 3:01 PM


Bravo!!
On Fri, 7 Apr 1995, Ted Alper wrote:
> Hey! I got it! Please bear with my ASCII diagrams. > > (for convenience, let's just worry about the CENTERS of the circles, > which lie in a 1 x 999 rectangle formed by lopping margins of length 1/2 > off the top, bottom, and sides. These centers must all be at least > 1 unit away from any other center. > > So, for example, in the square packing that gives 2000, the diagram is > > 1 2 3 4 5 6 ..996 997 998 999 1000 > ******....***** >   >   > ******....***** > > And in the supposedly more efficient packing (which really is > more efficient on the infinite plane) the boundary constraints only > give us 1999: > > 1 2 3 4 5 6 ..996 997 998 999 1000 > ******....***** > \ / > \ / > *****....**** > > But note there is some wiggle room in this! If the slanted sides are of > length 1, the HEIGHT is only sqrt(3/2) (remember, these are the > centers only  and they form equilateral triangles of side length 1). > > > That suggests the following shape, which takes advantage of the > entire height and thus enables us to pack a few more in: > > I can't do it to scale, but let me label the rows and give their > heights. Row D is on bottom, Row A is on top (so row A is height > 1 above row D). Row B is height sqrt(3)/2 above row D and row > C is height sqrt(3)/2 BELOW row A. > > So: to put it another way: Row B is 1  sqrt(3)/2 below row A, > Row C is sqrt(3)  1 below row B, Row D is 1  sqrt(3)/2 below row C. > > A ..4 6.. ..etc > ... \ / ... ... \ > B 2.. \ / ..8.. \ > / \ \ / / \ \ > C / \ ..5.. / \ ..etc > / \ ... ... / \ ... > D 13.. ..79.. > > > the connected lines are of length 1, so 1,2,3 is an equilateral > triangle, as is 4,5,6 and 7,8,9 etc. On the other hand, 2,3,4,5 > is a rhombus! > You may verify that triangle 3,5,7 is isosceles, with the > two identical sides of length 1. Since the altitude is > 1  sqrt(3)/2, we can use the pythagorean theorem to show > that the length from 3 to 7 is > 2 * sqrt( (1^2  (1  sqrt(3)/2)^2) ) > i.e. 2 * sqrt( sqrt(3)  3/4) > which is 1.981965 > > So the total length from 1 to 5 is 2.981965 > But this means that we may iterate the 1,2,3,4,5,6 diagram > 999/2.981965 times That's 335.01348 > > which means we get 335 copies of the 6 points, plus 1 more > (the very first point gets on in the 336ith iteration). > > THAT'S 2011. DONE! > >



