[psyang]

Spoiler!

Originally Posted by Mathgod

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.

Ok, I understand your argument. It's a bit moot anyways I think since you reduce the maximum area below.

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.

Ok, understood.

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.

Ok, I think I understand what you are saying. Yes, a diagram would be helpful.

I think you are saying: let's try to place 3 coins as far apart as possible without allowing a coin to fit between them. You start by placing 2 coins 2r apart (so centre distance is 4r). Draw an imaginary coin between them so the three coins are on the same horizontal line. Then you can place your third coin below the imaginary coin so that the coins form a T. The centre distance between this third coin and the first 2 coins is 2sqrt(2)r.

So my question is does it follow that the least efficient arrangement of coins is 2sqrt(2)rx by 2sqrt(2)ry? I don't know if that is obvious, since in my coin arrangement above, if I placed my first two coins slightly less than 2r apart, then the third coin could be moved much further away and still allow for no coin to rest in between them.