Puzzle for Today
Puzzle No. 272 – Thursday 19 July
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 try 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! A useful hint is to start packing coins from the middle not from one end.
Today’s #PuzzleForToday has been set by Mike Paterson, Emeritus professor in Computer Science at the University of Warwick
Click here for the answer
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? If you think you have a good answer to this, do let me know, but note that the really hard problem, still I think unsolved, is to prove the best optimum result.