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Topic:
how many coins?
Replies:
5
Last Post:
Apr 8, 1995 3:01 PM




Re: how many coins?
Posted:
Apr 7, 1995 11:53 PM


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!



