# Tiling a square with 3:1 rectangles

It is known that a square can be tiled with $$n$$ rectangles whose length is double their width for any $$n > 4$$. In particular, no two rectangles can overlap and no part of any rectangle is outside the square. The rectangles can be of different size.

I am now investigating the same problem for rectangles whose length is triple their width. Clearly if we can tile a square with $$m$$ rectangles then we can also tile it with $$m+4$$ rectangles using the "building up" method from the video. Also by placing the square into a $$2 \times 2$$ square grid (adding 3 additional squares), we can also tile it with $$m+9$$ rectangles. Using these two facts I was able to show that we can tile a square for any $$n > 26$$. However I still don't know which $$n \leq 26$$ do not have a tiling (other than the trivial ones). How can I find them?

I am also interested in the general version of this problem, ie., when the length of the rectangles is $$k$$ times their width.

Given a solution with $$m$$ rectangles, we can turn it into a solution with $$m+3$$ rectangles by subdividing one of the rectangles into four smaller copies of half the side length. We can also do $$m+4$$ via the method you described. Here is a construction that produces $$m+5$$: With these extensions and the base case $$n=3$$, we can get every integer except $$n=1,2,4,5$$; I haven't gone through a careful proof that all of these are impossible, but it's not too hard for $$4$$ and I suspect $$5$$ is doable with a bit of casework.

The general problem with $$1\times k$$ rectangles will have the trivial $$m=k$$ solution and the ability to extend solutions from $$m$$ to $$m+4$$ and $$m+n^2-1$$ for all $$n$$; this means that there will always be solutions past $$k+5$$. I expect each of $$k+1,k+2,k+5$$ could be tackled in generality, thereby solving the problem in all cases, but it might be fairly annoying to handle everything that comes up (especially in the $$k+5$$ case).

• Excellent answer, thank you! Jan 8, 2022 at 23:06
• Can you explain why there will be solutions for $m > k+5$ in the general case? Jan 9, 2022 at 0:00
• $n=5$ is not possible after all: puzzling.stackexchange.com/questions/114378/… Jan 9, 2022 at 7:30
• In the general case, we have a solution with $k$ rectangles, and we have the ability to enlarge any solution by 3 or 4 rectangles via the same constructions as in the $1\times3$ case. This means we can obtain $k,k+3,k+4,k+3+3,k+3+4,k+4+4,k+3+3+3,\ldots$ and so everything beyond $k+5$ is attainable. Jan 9, 2022 at 7:36
• Oh I see now. Thanks for the clarification. I was missing the fact that $m+3$ can be made for any $k$. Jan 9, 2022 at 9:03