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Say you had two coins, one was larger (twice the size) of the smaller coin.

Since the smaller coin was half the size, logically you would be able to store at least twice as many in the box. But since the spacing between smaller coins is less than larger coins (ie, less wasted space tesselating them), would you be able to store more than double?

Aequitas
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  • If the size of the box is large enough, the answer is yes. – Rushabh Mehta Jan 21 '20 at 00:34
  • Are the coins cylindrical? the same thickness? Is it twice the volume? area? diameter? – J. W. Tanner Jan 21 '20 at 00:37
  • @J.W.Tanner yes the coins are cylindrical, coins of the same type are the same thickness but the small and bigger ones are not necessarily the same (but can be if it allows a way to make the answer to the question true and false). So twice the volume but you can change any parameter(s) you consider to be 'size' (whether that is diameter/height/a mixture/etc) in order to get the double volume – Aequitas Jan 21 '20 at 00:42
  • Thinking about it; I think I am basically asking if, the percentage of wasted space when storing cylinders scales linearly with the size of the cylinder. So I guess imagining a single coin of diameter 5cm in a box 5cm wide, you would see that there's quite a lot of wasted space on the corners. So if you had a single layer of very tiny cylinders, obviously it would seem like there wasn't much wasted space, but there is lots more of them, so would the added up little spaces equal the wasted space with the big coin? I hope that makes sense, I'm not great with explaining haha. Thanks for your time – Aequitas Jan 21 '20 at 00:46
  • If you had coins that were half the diameter of the larger coins and, say, of the same thickness, then, assuming that they are neatly stacked in uniform raster rows, or as you say tessellated, then you could fit four times as many small coins in the same volume. I haven't considered what happens if alternate rows are offset, but I do know that there's a fixed area increase when you go from raster to hexagonal packing, so I expect the same result in that case as well. – Cye Waldman Jan 21 '20 at 23:57

1 Answers1

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Yes.

Proof:

Given a box of side length 1cm, you can fit zero coins of diameter 1.2cm but can fit at least one of the smaller coins at half the diameter (0.6cm).

Aequitas
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