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Re: how many coins?
Posted:
Apr 7, 1995 11:53 PM
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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
*------*------*------*------*------*....*------*------*------*------*
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*------*------*------*------*------*....*------*------*------*------*
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 1-------3.. ..7-------9..
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!
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