[math-fun] Tiling a square
Hi all, I know that 7 * 5^2 + 9 * 3^2 = 16^2, which suggests that 7 5x5 squares plus 9 3x3 squares can possibly tile a 16x16 square. Is there a relatively simple way to figure out if either: a) no, they won't, or b) yes, and here's how? Ideally, I'd like to learn the process, not just the answer, as I will probably have other combinations to consider. Thanks, Kerry Mitchell -- lkmitch@gmail.com www.kerrymitchellart.com
Each row (and each column) of any solution to the packing problem would give you a way to write 16 as a sum of 5s and 3s. Since there is only one such composition, 16=3+3+5+5, this could only work if the total area of the 3x3s were 6/16, and the total area of the 5x5s were 10/16, of the area of the 16x16. And it ain't. This kind of reasoning can only offer a proof of impossibility, of course. If more than one composition existed, you could get Diophantine constraints on how many rows of each type would need to exist, which might still show impossibility. Probably other people will have better ideas... --Michael On Wed, Feb 29, 2012 at 3:27 PM, Kerry Mitchell <lkmitch@gmail.com> wrote:
Hi all,
I know that 7 * 5^2 + 9 * 3^2 = 16^2, which suggests that 7 5x5 squares plus 9 3x3 squares can possibly tile a 16x16 square. Is there a relatively simple way to figure out if either: a) no, they won't, or b) yes, and here's how? Ideally, I'd like to learn the process, not just the answer, as I will probably have other combinations to consider.
Thanks, Kerry Mitchell -- lkmitch@gmail.com www.kerrymitchellart.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush.
Assertion: There is a tiling of a 16x16 square from 7 5x5 squares and 9 3x3 squares. Consider the topmost row of the 16x16 square. It contains a 1-unit "slice" of each of four of the smaller squares; from left to right this must be 5+5+3+3, 5+3+3+5, 5+3+5+3, 3+5+5+3, or a reflection of one of these. Now consider the 4th row. At this point the 3x3 squares have ended but we're still in the 4th "slice" of the original two 5x5 squares. The only way to fit new squares in without leaving impossibly small gaps is to add two more 3x3 squares in the same horizontal positions as before. Thus the same pattern (5+5+3+3, 5+3+5+3 or whatever) continues. In a similar manner, when we get to the 6th row we are forced again to place two more 5x5 squares in the same positions. This continues all the way up to the 15th row, after which we have placed 6 5x5 squares and 10 3x3 squares. But we meant to use only 9 3x3 squares, therefore the original assertion must be false. - Robert Munafo On Wed, Feb 29, 2012 at 15:27, Kerry Mitchell <lkmitch@gmail.com> wrote:
Hi all,
I know that 7 * 5^2 + 9 * 3^2 = 16^2, which suggests that 7 5x5 squares plus 9 3x3 squares can possibly tile a 16x16 square. Is there a relatively simple way to figure out if either: a) no, they won't, or b) yes, and here's how? Ideally, I'd like to learn the process, not just the answer, as I will probably have other combinations to consider.
Thanks, Kerry Mitchell
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
participants (3)
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Kerry Mitchell -
Michael Kleber -
Robert Munafo