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This question already has an answer here:

I recently heard about this intriguing puzzle:

I have a tray of length 5 and width 2 so 10 round coins of width 1 will fit in it snugly without overlaps. No room for another. Similarly, a tray of length 50 will accommodate only 100 coins. Things get more interesting with a longer tray! A tray of length 500 and width 2 can accommodate at least 1001 coins. Show how this can be done.

To be clear, this is just about fitting non-overlapping circles in rectangles, so no trickery with funny-shaped coins or thermal expansion coefficients!

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marked as duplicate by Riley, Alconja, Chowzen, JMP, w l Jul 24 '18 at 5:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Explanation

The diagram below shows how you could do it. Diagram shows the right half of the tray with 125 coins (1..125) aligned to the top of the tray. Each is in contact with its horizontal neighbors (blue lines show contact). There are a further 125 coins (126..250) in the bottom tray. Each is in contact with its horizontal neighbors (green lines), but the even numbered coins are in contact with a neighbor above (red lines), and the odd numbered coins are in contact with the bottom of the tray. Moving to the right, each connecting red line is slightly more vertical, until at the 64th pair (coins 125 and 250) it is vertical, so coin 250 just fits in the tray. Below and to the left of coins 1 and 126, there is space for half a coin, so by reflection at the dotted line we could fit 2*(125+125+1/2)=501 coins in the tray.

Figure

enter image description here

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  • $\begingroup$ How do you know you can achieve this by 250 coins? Perhaps it takes more or (I think) quite a few less. $\endgroup$ – Trenin Jul 24 '18 at 18:13
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Consider a very long tray and lay a row of touching coins along one long edge. Now lay a row of coins along the opposite long edge but shifted along by half a coin width. (I wish I could show a diagram here, but you might like to get out some coins to play with.) Notice that now there is clear space between the two rows of coins. This allows us to make a slight zigzag in each, and any such wiggle shortens the length of each row. For a long enough tray this must eventually make enough space for an extra coin or so.

For a more mathematical challenge:

what is the smallest n so that 2n+1 circles of diameter 1 can be packed without overlap in an n by 2 rectangle?

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  • $\begingroup$ I think you just copied and pasted your answer! $\endgroup$ – Anush Jul 24 '18 at 4:48
  • $\begingroup$ @Anush hahaha.. I hope you did same think for your question $\endgroup$ – CR241 Jul 24 '18 at 4:51
  • $\begingroup$ Questions and answers are not symmetric :) $\endgroup$ – Anush Jul 24 '18 at 4:52

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