# Five 3:1 rectangles tiling a square

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.

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

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

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!

• This is excellent. Thank you! Jan 9 at 9:12