[psyang]

I think it's a great start but not a valid solution. Comments/questions below in blue:

Originally Posted by Mathgod

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Spoiler!

Let r be the radius of the coin. The dimensions of the table can be larger than x coins wide and y coins long, so the area of the table is larger than x*2r*y*2r = 4xyr^2

However, since the coins are close enough to prevent any more coins from fitting between them, the total area must be smaller than 2x*2r*2y*2r = 16xyr^2

Can you clarify how you get 2x*2r*2y*2r as a bounding area for the table? I'm not seeing where that is coming from.

Furthermore, the coins must also be close enough to prevent any coin from fitting in the two-dimensional spaces between coins on the table. As a consequence, the centres of the coins must not be farther apart than 2sqrt(2)r.

Here, you are assuming the coins are packed in a grid, but that's not necessarily the case. What if you created a grid but added an infinitely small space between touching coins (ie. space between two coins in x direction and two coins in y direction). You still couldn't place a coin in the space in between the 4 coins, but the centers of the diagonal coins would be more than 2sqrt(2)r apart.


This means that the total area of the table must be smaller than x*2sqrt(2)r*y*2sqrt(2)r = 8xyr^2

Now place coins, starting at the top left corner, exactly 2r apart, in a grid pattern, then a 2nd layer precisely positioned to cover the gaps left by the first layer. All area is covered, and the non-overlapping area that each coin contributes is r^2/2*4 = 2r^2. Since there are 4xy total coins, the total area covered is 8xyr^2

The solution isn't quite that simple, as it doesn't quite explain what happens on the sides and on the corners, but that's the gist of it.