[Mathgod]

Spoiler!

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.

x is the number of coins going horizontally, 2r is the diameter of the coin, then I multiply by 2 to take into account not just the length contributed by the coins themselves, but also the length contributed by the spaces between the coins.

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.

I think you may have misinterpreted what I'm saying here. I didn't say anything about a "packed" grid or the coins touching. I'm talking about moving the coins away from each other horizontally and vertically, to the maximum amount possible without creating enough space to place a coin the created space. This means a diagonal hypotenuse approaching 2r, which means the horizontal/vertical side length becomes sqrt(2)r. This is half the horizontal (or vertical) distance between the center of one coin and the center of the next coin. Hence, the horizontal distance from coin center to coin center approaches 2sqrt(2)r, but can't actually reach that distance, otherwise you can now fit a coin in the space created.

Admittedly, it may be hard to visualize what I'm saying without me actually providing diagrams.