If you divide a square into N similar rectangles, what proportions can these rectangles have?
A while back, Ian Henderson (@ianh) tackled this problem for N = 7. That's right: SEVEN similar rectangles! Nobody else got that far.
He found 1371 options for what the proportions of these rectangles could be. The picture here is his drawing of all the options.
Most of us congratulated him and went about our daily business. But now David Gerbet has checked the calculation - and he seems to have found one more option!
After some work, he figured out which of his 1372 options was not on Ian Henderson's list. So it would be great if someone could check this. I guess anyone could check to see if this solution really is a solution... but I expect only Ian can figure out whether and why he didn't find it before.
I wrote a blog article with more details:
https://johncarlosbaez.wordpress.com/2023/03/06/dividing-a-square-into-7-similar-rectangles/
And I'll show you Gerbet's new claimed solution on the next page here.
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Here is David Gerbet's new solution! He writes:
"After running the computation again, I was able to identify the possibly missing guy exactly: the 1055th ratio. The pictures of the partitions look all the same for smaller and larger ratios in the tables from Ian Henderson and the one I computed. You find it at page 106 in my table of partitions with 7 rectangles.
The ratio is the only real root of the polynomial
x⁷−3x⁶+9x⁵−10x⁴+12x³−7x²+2x−1
You can find a picture of a partition with this ratio in the files attached. Note that Ian Henderson has computed partitions with other ratios, but the same topology as this one, e.g. the 1022th ratio."
Here is the table that David is talking about:
This apparent new solution is made from rectangles with proportions of 0.7148589 to 1. But is it real? Should we welcome it to the elite club?
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@johncarlosbaez Without checking the polynomial: it seems to me it would be a coincidence if it was real. There appears to be more than one constraint on the ratio.
@johncarlosbaez @Lisanne Indeed, it isn’t obvious to me in this case. For many topologies & rectangle orientations, it’s intuitively clear that you could smoothly vary the ratio, and find the point where it a) forms a square, and b) all the pieces fit. Not here. In this case, my first impression was that the horizontal position of the split line going up from the pink subrectangle is a separate variable—one must make the two sides match, both top and bottom, and the pink block must have the correct ratio as well…
I’ll take a look at Lisanne’s proof. Thanks!
@rgs - I'm not sure what you mean, but each solution to this sort of problem can be described as the solution of a single polynomial equation. That's not obvious, but it was proved by @Lisanne here earlier.