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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 5:01 PM


I can't draw with dots as well as Jim, but suppose we take his configuration. Alas, I cleverly deleted the problem. If I don't have it right, I think the same construction can be used to explore the correct parameters.) I'm taking that to be : Coins with a diameter of one cm A band 2 cm high and very long (1000 cm?) Two coins on top of each other, flush left, followed by a coin whose center is 1 cm high, wedged next to the other two, followed by two coins on top of each other, as close as possible to the third, etc.
Consider just the first three coins; Let A and B be the ones on top of each other (A on top) and C be the third. Draw the equilater triangle connecting their centers. Now drop a vertical line from the center of C to its circumference (point x). Point x, together with the centers of B and C form a right triangle, with the right angle at x. The hypotenuse of the triangle is 1 cm; it is a 306090 triangle with the 30 degree angle at the center of B. So, the line from the center of B to x is sin 60 = .8660.
Now consider the rectangle formed by the upper, lower and left sides of the band, and the line from x through the center of C, extended to the top of the band. This rectangle contains 2.5 circles. It uses up 0.5 + .866 = 1.366 of the length of the band. The next 1.366 of the band also contains 2.5 circles, and so on.
Therefore, this packing method yields a density of 2.5 circles/1.366 linear cm, or 1.830 ci/cm. By contrast, the obvious packing, 2 above 2, next to 2 above 2, etc, has 2 circles every cm. This would suggest that Jim's construction does not pack coins with sufficient density to meet the constraints.
===============================
Jim:
Can you prove this?
The area of the rectangle is 2000 cm^2, and the circle is pi/4 cm^2. So, if the coing was "soft", we can fit 2000 / (pi/4) ~ 2500. The question is, off setting the coins will really reduce the waste.
By the way, the name Graham sounds familiar. Does anyone have a reference?
************************************************************************ * Tad Watanabe email * * Dept. of Mathematics watanabet@toe.towson.edu * * Towson State University watanabet@towsonvx.bitnet * * Towson, MD 21204 tad@midget.towson.edu * * (410) 830  3585 (410) 830  4149 FAX * ************************************************************************
On Fri, 7 Apr 1995, Jim Osborn wrote:
> > There are three rows of coins instead of two > > .. .. > . . . . > . . .. . . > .. . . .. > .. . . .. > . . .. . . > . . . . > .. .. > > There are more coins packed into the middle row then are lost by the > missing pairs of coins. >



