Some people on Mathstodon put a lot of work into this and made some nice progress. But there’s been a surprising new twist! I’m not talking about how the answer to this puzzle is now listed on the listed on the Online Encyclopedia of Integer Sequences as sequence A359146. I’m not even talking about the fact that the New York Times ran an article about this puzzle:

• Siobhan Roberts, The quest to find rectangles in a square, New York Times, February 7, 2023. Open-access version here.

]]>People have raised this issue on Mathstodon but not really worked on it.

I think there’s a huge amount of interesting structure in the case of dividing a square into similar rectangles: see my blog article

• Guillotine partitions and the Hipparchus operad.

and also the comments on the original problem starting here—especially the comments by David Speyer and Jeffery Opoku.

Unfortunately I haven’t had time to work on these things in the last month.

]]>Yes it does!

]]>Thanks. I’m trying to understand this. Does mean “maybe we include a factor of maybe not?”

]]>Let be the long side of rectangle and the short side, so the ratio is . The initial square will have side length s.

Now for each vertical line segment (always take the full segment) in the rectangulation we get a relation

For the outer line segments (the sides of the square) do something similar with

And similarly for the horizontal segments.

Using some induction argument you can see that this always gives you relations. (This is easy for guillotine cuts a little bit harder in general). Here you need the generic property.

However the relation on the right side is implied by all the other vertical relations and the same is true for the bottom side and the horizontal relations. Removing these we end up with relations.

Some intense staring will show that a solution to these equations is not just necessary but also sufficient to get a similar-rectangulation.

Now we will have a closer look at the relations. We have variables ( and ). So we can fit them into an by matrix and we need to solve the equation . Now this has a (non-zero) solution if and only if and this is just a polynomial in . Since the column corresponding to the variable has only in it this polynomial can have degree at most .

This proof also suggests an efficient way to compute and . Write down the matrix, solve the determinant polynomial. Substitute this back in and solve a linear equation. You should be carefull that even if there is a solution some of the might be negative. Geometrically this corresponds to two rectangles overlapping, but the overlap also having the same ratios!

Bonus:

We can also estimate the coefficients of . There is a naive bound coming from the Leibniz formula saying that the sum of all the coefficients is at most Can we do better?

Right!

]]>I’d love to see the proof! And I’d love to see that matrix. I’m working on this stuff myself, slowly.

]]>Thanks very much! That paper is available here.

(Elsevier may be offering it to me for free because they’re spying on me and know I’m at U. C. Riverside, even though I deleted all their cookies a while back. I read a great blog article about this topic recently but now I can’t find it.)

]]>I can type out a proof for this later if someone wants to see it. In fact the polynomial is the determinant of a matrix with only +-1,+-X,0 in it.

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