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Can you fully tile a square with 5 rectangles such that:

  • Every rectangle has 3:1 ratio, ie., their length is triple their width.
  • No part of any rectangle is outside the square.
  • No two rectangles overlap.

Note that rectangles can have different size.

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  • $\begingroup$ I don't know the answer to this puzzle. $\endgroup$ Jan 9 at 0:05

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From this MSE question we know that all tilings of any rectangle by 5 rectangles – except one – may be built up by joining tilings by a smaller number of rectangles edge-to-edge. Without loss of generality let their aspect ratios be $a,b$, then if they are joined on the "1" ends the combined tiling has aspect ratio $a+b$. Note that once we have a ratio-$a$ rectangle we also have a ratio-$1/a$ one, so we need to take union with reciprocals of the aspect ratios found at each stage.

What we find from this recursive process is

it's not possible, and the closest you can get is an aspect ratio of $\frac{57}{56}$.

Now

to handle the last remaining case – the "prime" tiling in the MSE question – let the square have side 1 and the NW/SW/SE/NE rectangles have lengths $s_1a,s_2b,s_3c,s_4d$ and heights $t_1a,t_2b,t_3c,t_4d$ respectively, where $\{s_i,t_i\}=\{1,3\}$. Solving this linear system for each of the 16 possibilities yields no solution in 6 cases, a negative value for one of $a,b,c,d$ in 8 cases and a solution which leaves the central rectangle a perfect square, not a $3:1$ rectangle in the last 2.

In conclusion

there is no solution!

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  • $\begingroup$ This is excellent. Thank you! $\endgroup$ Jan 9 at 9:12

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