## December 22, 2022

### Dividing a Square into Similar Rectangles

#### Posted by John Baez

If you divide a square into some fixed number of similar rectangles, what proportions can these rectangles have? We’ve been having fun thinking about this on Mathstodon, and here is a progress report.

If you divide a square into 3 similar rectangles, what proportions can these rectangles have? There are three options. The third is much more complicated than the first two:

• We can divide the square into three rectangles that are 1/3 as long in one direction as the other, as in the first picture.

• We can divide it into three rectangles that are 2/3 as long in one direction as the other, as in the second picture.

• We can divide it into three rectangles that are $x$ times as long in one direction as the other, as in the third picture.

What’s $x$? The yellow rectangle has height $1$ and width $x$, so the blue rectangle has width $1-x$ and height $x(1-x)$, while the red one has width $1-x$ and height $(1-x)/x$. The heights of the blue and red rectangles must sum to $1$, so

$x(1-x) + (1-x)/x = 1$

or

$x^2 (1-x) + (1-x) = x$

$x^2 - x^3 + 1 - x = x$

$x^3 - x^2 + 2x - 1 = 0$

and the solution of this cubic is

$\displaystyle{ \frac{1}{3} \left(1 - 5 \left(\frac{2}{11 + 3 \sqrt{69}}\right)^{1/3} + \left(\frac{1}{2} \left(11 + 3 \sqrt{69}\right)\right)^{1/3}\right) }$

so

$x \approx 0.56984$

The reciprocal of this number is the square of a famous constant: the plastic ratio, $\rho$. This is like a cheap imitation of the golden ratio, with

$\rho^3 = \rho + 1$

Why is the third option so much more complicated than the other two? As we’ll see, it’s because the first two have all the rectangles ‘pointing the same way’, while the third does not: it has a mix of rectangles ‘standing up’ and ‘lying down’.

To see the pattern, it helps to do a harder problem.

If you divide a square into 4 similar rectangles, what proportions can these rectangles have? A lot of people on Mathstodon worked on this puzzle! Here Dan Piker lists the 11 options we’ve found:

Note: there are more than 11 ways to divide a square into 4 similar rectangles, since you can rotate and reflect these pictures, and also rearrange the rectangles in some of them. But we’ve only found 11 possible proportions for the rectangles. Stefano Gogioso has sketched a proof that these are all the options, and Rahul Narain has given a computerized proof that these are all the options obtained from ‘guillotine cuts’:

A ‘guillotine cut’ is a straight line going from one edge of an existing polygon to the opposite edge. And I believe there’s no way to dissect a square into 4 rectangles that doesn’t use guillotine cuts.

In the process of listing the options, we discovered something interesting: the rectangles have rational proportions only in options 1, 4, 7, 10 and 11 here:

And these are precisely the options where all the rectangles are pointing the same way! (They happen to all be lying down, wider than they are tall. But of course they’d all be standing up if we rotated the pictures.)

Puzzle 1. Show that if you subdivide a square into $n$ similar rectangles that are all pointing the same way, the ratio $x$ of their short side to their long side must be a rational number.

Puzzle 2. Show that the converse is not true.

We also discovered another pattern, too: when $x$ is not rational, it is the solution to a cubic equation with integer coefficients.

Does this other pattern persist when we subdivide a square into 5 similar rectangles? No, alas! Ian Henderson and Rahul Narain seem to have shown that exactly 51 proportions are possible when we subdivide a square into 5 similar rectangles. Henderson drew them:

while Narain listed the polynomial equations obeyed by the 51 possible proportions. Some obey a quartic equation with integer coefficients, not a cubic.

To be honest, Narain only considered ways of subdividing a square into 5 similar rectangles using guillotine cuts. But this should be okay, since I believe there’s just one way to subdivide a square into 5 rectangles that cannot be done using guillotine cuts. It’s shown here in a paper by Robert Dawson, in fact:

However, when we try to make all 5 rectangles similar, the central one shrinks to a point.

Ian Henderson also believes that when you subdivide a square into 6 similar rectangles, 245 proportions of rectangles are possible:

And for subdivisions of a square into 7 similar rectangles, he believes 1371 proportions are possible:

You can click to enlarge these last two pictures.

However, he adds, “I’m not 100% confident in these numbers.” So, someone should check his work.

Puzzle 3. Show that you can divide a square into 6 similar rectangles in a way that cannot be done via guillotine cuts.

Okay, back to something simpler: what proportions are possible when we divide a square into 4 similar rectangles? Let’s work it out!

For each option we get a number $0 < x \le 1$ describing the proportion of the rectangles that subdivide the square. This number is the length of the short side divided by the length of the long side. The 11 options below are listed from the smallest possible value of $x$ to the largest.

1) Option 1 is to divide the $1 \times 1$ square into 4 rectangles that are 1/4 as tall as they are wide. So, the number we get from this option is 1/4.

Note that if we rotated this option we’d get tall skinny rectangles, but the number would still be 1/4.

2) In option 2, the bottom two rectangles have width 1 and height $x$. Thus the top two have height $1-2x$.

Since the green rectangle is $x$ times as tall as it is wide, its width must be $(1-2x)/x$. The yellow rectangle thus has width

 1 - (1-2x)/x = 1 - x^{-1} + 2 = 3 - x^{-1}

Its height is this divided by $x$, namely $3x^{-1} - x^{-2}$. But we know its height is $1-2x$, so

$1-2x = 3x^{-1} - x^{-2}$

or

$2x^3 - x^2 + 3x - 1 = 0$

so

$x \approx 0.34563$

3) For option 3 the red rectangle has height $x$, so the blue one has height $1-x$ and width $x(1-x) = x - x^2$, so the other two have width $1-(x - x^2)$ = $1-x+x^2$ and height $x - x^2 + x^3$. The total height of red, green and yellow is $1$ so $x + 2(x - x^2 +x^3) = 1$ or

$2x^3 - 2x^2 + 3x - 1 = 0$

This gives

$x \approx 0.39661$

4) Option 4 is nice: it has left-right symmetry, and its rectangles can be rearranged to give another option with left-right symmetry.

The red and blue rectangles have height $x$. The green and yellow ones thus have height $1-2x$, and thus width $(1-2x)/x$. But by symmetry we know they must have width 1/2, so

$(1-2x)/x = 1/2$

or

$1-2x = x/2$

or

$1 = 5x/2$

or

$x = 2/5 = 0.4$

Not a cubic equation this time—linear! So x is rational.

Notice that options 3, 5, 6, and 7 are all topologically the same. They were analyzed by Lisanne, and her work helped me a lot.

5) In option 5 the red rectangle has height $x$, so the blue has height $1-x$ and thus width $(1-x)/x$. The yellow and green thus have width $1 - (1-x)/x = 2 - x^{-1}$, hence height $(2 - x^{-1})/x = 2x^{-1} - x^{-2}$.

The heights of yellow, green and red must sum to 1:

$2(2x^{-1} - x^{-2}) + x = 1$

so we get a cubic:

$x^3 - x^2 + 4x - 2= 0$

and the solution is

$x \approx 0.53318$

6) In option 6 the red rectangle again has height $x$, so the blue again has height $1-x$ and width $(1-x)/x$.

The yellow and green again have width $1 - (1-x)/x = 2 - x^{-1}$, but now they’re different: the yellow has height $x(2 - x^{-1})$ while the blue has height $(2 - x^{-1})/x$.

The heights of yellow, green and red sum to $1$:

$x(2 - x^{-1}) + (2 - x^{-1})/x + x = 1$

so

$3x^3 - 2x^2 + 2x - 1 = 0$

or

$x \approx 0.55232$

7) In option 7, like options 5 and 6, the red rectangle has height $x$, so the blue has height $1-x$ and width $(1-x)/x$.

Thus, the yellow and green again have width $1 - (1-x)/x = 2 - x^{-1}$. But this time both have height $x(2 - x^{-1})$.

Yet again, the heights of yellow, green and red sum to 1:

$x + 2x(2 - x^{-1}) = 1$

so

$5x = 3$

and

$x = 3/5 = 0.6$

The equation was linear, so $x$ is rational!

Options 8, 9, and 10 are also all topologically the same.

8) In option 8 the red rectangle has height $x$ so the blue has height $1-x$ and width $(1-x)/x$. The green and yellow also have height $1-x$, but width $x(1-x)$.

The widths of blue, green and yellow sum to 1:

$(1-x)/x + 2x(1-x) = 1$

so

$2x^3 - 2x^2 + 2x - 1 = 0$

and

$x \approx 0.64780$

9) In option 9 the red rectangle has height $x$ so the blue and green have height $1-x$ and width $(1-x)/x$. The yellow also has height $1-x$, but width $x(1-x)$.

The widths of blue, green and yellow sum to 1:

$2(1-x)/x + x(1-x) = 1$

so

$2x^{-1} - 2 + x - x^2 = 1$

or

$x^3 - x^2 + 3x - 2 = 0$

so

$x \approx 0.71523$

10) Option 10 is more symmetrical than 8 or 9 since all three rectangles on top are congruent.

The red rectangle has height $x$ so the blue, green and yellow all have height $1-x$ and thus width $(1-x)/x$.

The widths of blue, green and yellow sum to 1:

$3(1-x)/x = 1$

so

$3(1-x) = x$

or

$3 = 4x$

or

$x = 3/4 = 0.75$

Another linear equation, with a rational solution!

11) Finally, option 11 is the second one where all four rectangles are congruent. This time they’re squares! Clearly

$x = 1$

So, we see in these examples that $x$ is rational precisely when all rectangles are ‘pointing the same way’: in the way Dan drew them, they’re all at least as wide as they are tall.

I’m not quite sure where to go with this research next. But I think the connection to double categories, hinted at in this paper with the picture of the pinwheel configuration, is interesting:

Maybe we should seriously study the double category where 2-cells are rectangles that are subdivided into smaller rectangles by guillotine cuts!

But I also keep hoping there’s some interesting number-theoretic significance to the proportions that come up when we divide a square into the similar rectangles. Can anyone see interesting patterns in Rahul Narain’s table of the polynomials obeyed by these ratios when we divide a square into 5 similar rectangles? It might be easiest to start by focusing on the rational numbers. His $u$ is my $1/x$:

u - 5
u^3 - 4*u^2 + u - 3
u^3 - 4*u^2 + 2*u - 4
2*u - 7
u^3 - 4*u^2 + 3*u - 3
2*u^3 - 6*u^2 + u - 2
u^3 - 3*u^2 + 2*u - 5
u^3 - 3*u^2 + 2*u - 4
2*u^3 - 6*u^2 + 2*u - 2
u^3 - 3*u^2 + u - 1
3*u - 8
u^3 - 3*u^2 + 3*u - 5
2*u^3 - 5*u^2 + u - 2
u^5 - 3*u^4 + 3*u^3 - 4*u^2 + u - 1
-2*u^2 + 6*u - 3
2*u^3 - 5*u^2 + 2*u - 3
3*u - 7
u^4 - 3*u^3 + 3*u^2 - 4*u + 2
2*u^3 - 5*u^2 + 3*u - 3
u^3 - 3*u^2 + 4*u - 4
-u^2 + 4*u - 4
4*u - 8
3*u^3 - 6*u^2 + u - 1
2*u^3 - 4*u^2 + 2*u - 3
u^3 - 2*u^2 + 3*u - 5
3*u^3 - 6*u^2 + 2*u - 2
u^5 - 2*u^4 + 3*u^3 - 5*u^2 + u - 1
2*u^3 - 4*u^2 + 3*u - 4
4*u - 7
u^3 - 2*u^2 + 4*u - 6
-u^3 + 3*u^2 - 4*u + 3
2*u^3 - 5*u^2 + 4*u - 2
2*u^3 - 4*u^2 + 3*u - 3
3*u^3 - 5*u^2 + 2*u - 3
5*u - 8
u^5 - 2*u^4 + 3*u^3 - 4*u^2 + u - 1
-3*u^2 + 6*u - 2
2*u^4 - 4*u^3 + 3*u^2 - 3*u + 1
3*u^3 - 5*u^2 + 2*u - 2
4*u^3 - 6*u^2 + u - 1
-2*u^3 + 4*u^2 - 3*u + 2
2*u^3 - 3*u^2 + 3*u - 4
u^5 - 2*u^4 + 3*u^3 - 4*u^2 + 2*u - 1
5*u - 7
u^3 - 2*u^2 + 3*u - 3
2*u^3 - 3*u^2 + 2*u - 2
-u^3 + 2*u^2 - 4*u + 4
3*u^3 - 5*u^2 + 3*u - 2
3*u^3 - 4*u^2 + u - 1
4*u^3 - 6*u^2 + 2*u - 1
4*u - 5
2*u^3 - 3*u^2 + 4*u - 4
-5*u + 6
6*u - 7

Posted at December 22, 2022 6:15 PM UTC

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### GARTH

What if instead of a square, you require the outer rectangle to have the same aspect ratio as the pieces?

The results for 2 and 3 pieces have similar arrangements to the square ones (I suspect this will generally be the case), but of course the numbers are different: sqrt(2)/2, sqrt(3)/3, (-1+sqrt(17))/4, and a cubic root approximately equal to 0.72449.

The first rational number (1/2) doesn’t show up until 4 pieces.

Posted by: Garth Rose on December 23, 2022 6:50 AM | Permalink | Reply to this

### Re: GARTH

Rectangle dissection into similar rectangles.
https://math.stackexchange.com/questions/2709153/rectangle-dissection-into-similar-rectangles

Posted by: Ed Pegg on December 28, 2022 6:20 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

It would be good to pair each rectangle with its polynomial. After that, it would be good to pair each with the polynomial’s number field discriminant. For example, -31 -> {3 - 4 u + 3 u^2 - u^3, -1 + u - 4 u^2 + 3 u^3} are both in the space of the supergolden ratio. -44 -> {4 - 4 u + 2 u^2 - u^3} is in the space of the tribonacci constant. -76 -> {-2 + 2 u - 6 u^2 + 2 u^3, -1 + u - 3 u^2 + u^3} is in the space of x^3=2(x+1). the value of x^24 -24 is 885479.77799 and the value of E^(Pi Sqrt[19) is 885479.77768. The same root as the Heilbronn Square for 12 points or the Melissen 12 disk covering.

Posted by: Ed Pegg Jr on December 28, 2022 6:17 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

I don’t know what you mean by “pair each rectangle with its polynomial”. What has a polynomial, as far as I know, is a way of dissecting a square into $N$ similar rectangles.

What are your numbers like -31, -44, etc?

It sounds like you’re doing something really interesting, but I don’t understand it.

Posted by: John Baez on December 28, 2022 11:19 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

I wrote a blog about some of these special values at “Shattering the Plane”. For the polynomials. one of the first you discuss is the polynomial for the plastic ratio, ρ^3 = ρ+1. The number field discriminant for this polynomial is -23. Once you know the discriminant, you can explore the number field. At http://www.lmfdb.org/NumberField/ you can put in a discriminant, for example. A nice related problem is to divide a rectangle into similar rectangles.

Posted by: Ed Pegg Jr on January 23, 2023 10:10 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

I did dissections of squares into 4 and 5 similar rectangles a few years ago, and started on dissections into 6 similar rectangles before setting the project aside to work on something else… Nice to see somebody finish the job!

Something I noticed is that all of the polynomials for these kinds of constructions seem to have alternating signs. I had some informal reasoning about why that might be the case, but would love to see a formal proof.

Posted by: Andrew Hudson on December 31, 2022 6:03 AM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

Wow, thanks for that observation about alternating signs. I’ve been blind!

I’m going to post this as a challenge on Mathstodon.

Conjecture 1: the ratio of the long side to the short side of the rectangles in a dissection of a square into $N$ similar rectangles is a solution of a polynomial equation with integer coefficients of degree at most $N$.

Conjecture 2: we can choose this polynomial to have alternating coefficients.

Posted by: John Baez on December 31, 2022 9:58 AM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

Relevant to conjecture 2 is the following fact:

Let f(x) be a polynomial with real coefficients. Then f(x) divides a polynomial g(x) with alternating coefficients if and only f has no negative real roots. Moreover, if f has rational coefficients, we can arrange that g does as well.

This result is usually stated instead asking for f to have no positive real roots and asking that g have positive coefficients, which is obviously equivalent. See https://math.stackexchange.com/questions/1547305

So, if we are given a positive real algebraic number alpha, and if none of the Galois conjugates of alpha are negative, then alpha is the root of a polynomial with alternating coefficients.

Posted by: David Speyer on January 4, 2023 6:13 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

If we take a “ground up” approach instead, one starts with a rectangle with aspect ratio $s$, and asks what ratios can be constructed by joining such rectangles together.

If we consider only constructions that arise from guillotine cuts, then I believe they are “generated” by using the following operations:

1. Reflection: Taking a previous construction and rotating it 90 degrees.

2. Join: Taking two previous constructions scaled to share a common height and joining them on their vertical sides.

It follows that the set of constructible ratios from $s$ is a subset of $(0, \infty)$ closed under multiplicative inverses (reflection) and addition (joining). In fact, it is the closure of $\{s\}$ under these two operations.

In general, given a field $F$, let us call a subset $S \subseteq F$ closed under addition and multiplicative inverses a slice(just making up names here, feel free to not use this) of $F$. Given an element $s \in F$, we shall let $\langle s \rangle$ be the slice generated by $x$.

Then our question about tiling with guillotine cuts becomes: for what $s \gt 0$ does $1 \in \langle s \rangle$?

A few remarks:

1. By closure, one has that $\mathbb{Q}^{+} s \subseteq \langle s \rangle$. It follows that all positive rational numbers are achievable from guillotine cuts.

2. Similarly, by closure, it is equivalent to ask for $s$ such that $\mathbb{Q}^{+} \subseteq \langle s \rangle$.

3. To each “equivalence class”(very roughly speaking, this is topological equivalence + orientation information - rearrangement of rectangles) of guillotine diagram, we may associate a rational expression. Specifically, let $\operatorname{Frac}(\mathbb{Z}[x])$ be the field of integer rational expressions. Then there exists an expression in $\langle x \rangle$ corresponding to the diagram class. For example, the third diagram in your post corresponds to the expression

$x + (x + x^{-1})^{-1} = x + \frac{1}{x + \frac{1}{x}} = \frac{x^3 + 2x}{x^2 + 1}.$

With this said, we may proceed in proving* “Conjecture 2”, namely that the associated polynomial to a diagram class has alternating integer coefficients. It is enough to prove that the associated rational expression has a odd/even numerator with positive integer coefficients and an even/odd denominator with positive integer coefficients. Note that this is the case for the rational expression shown earlier, which produces the associated polynomial $(x^3 + 2x) - (x^2 + 1)$ (in general, the associated polynomial given an expression $p(x)/q(x)$ is $p(x) - q(x)$).

Clearly, $x = x/1$ is of this form. Now, we show that this form is preserved under the operations generating the slice. Clearly the coefficients of the numerator and denominator remain positive integers under inversion. One may easily verify this remains the case for addition of rational expressions.

Inversion flips the parity of the numerator and denomiator; for addition one has: $\frac{\text{even}}{\text{odd}} + \frac{\text{even}}{\text{odd}} = \frac{\text{even} \cdot \text{odd} + \text{even} \cdot {odd}}{\text{odd}\cdot\text{odd}} = \frac{\text{odd}}{\text{even}},$ $\frac{\text{odd}}{\text{even}} + \frac{\text{even}}{\text{odd}} = \frac{\text{odd} \cdot \text{odd} + \text{even} \cdot {even}}{\text{even}\cdot\text{odd}} = \frac{\text{even}}{\text{odd}}.$

Thus all rational expressions in $\langle x \rangle$ can be written in this form, proving* the conjecture.

Note: There are expressions equivalent to rational expression associated to a diagram, many of which may not have even/odd numerator or denominator. However, I think this is fine, as the associated expression should be obtained in the most “straightforward” way from an expression such as $x + (x + x^{-1})^{-1}$ by only setting common denominators.

Question (potentially easy to answer): Are the numerator and denominator of the rational expression associated to a diagram always coprime?

Posted by: Jeffery Opoku on January 1, 2023 10:45 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

Nice analysis! We can also use this to analyze the restricted case where all rectangles are required to have the same orientation. (As in John’s Puzzle 1 above.) In that case, the formula is always of the form qs or q/s for q rational. The operations are

(1) Add together a bunch of $q_i s$ terms to make $(\sum q_i) s$.

(2) Add together a bunch of $q_i/s$ terms to make $(\sum q_i)/s$.

(3) Take the reciprocal, to turn $qs$ into $1/(qs)$ and vice versa.

One might wonder whether every positive rational number can still occur, but the answer is yes: Write s as an Egyptian fraction $1/a_1+1/a_2+...+1/a_k$. Then divide your square into k columns, and divide the j-th column into $a_j$ rectangles. It might be interesting to investigate the smallest number of rectangles required to achieve a given rational number.

Posted by: David Speyer on January 4, 2023 7:05 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

Jeffrey’s post above raises the very natural question: What is the free slice on s? In other words, inside the field Q(s), what is the slice generated by s.

I thought I had an answer to this question, but it fell apart when I tried to write it up. So I’ll just pose it as a question. In particular, here is an example which I can’t resolve:

Is (s^3+s)/(s^2+2) in the slice generated by s?

Posted by: David Speyer on January 4, 2023 7:59 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

Here is a peculiar property of functions in the free slice: Let $0 \leq \theta \leq \pi/2$ and let $W(\theta) = \{ z \in \mathbb{C} : -\theta \leq \text{arg}(z) \leq \theta \}$. (The letter $W$ stands for “wedge”.)

Any function in the free slice on $s$ maps $W(\theta)$ into itself. Proof: $W(\theta)$ is closed under addition and inversion.

In particular, this means that $(s^3+s)/(s^2+2)$ is NOT in the free slice; it does not map the right half plane to itself.

Posted by: David Speyer on January 4, 2023 9:29 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

More thoughts: This means that all roots of a(s) and b(s) must be negative reals. Since the coefficients of a(s) and b(s) are nonnegative, they can’t have positive real roots.

Suppose for the sake of contradiction that z is a nonnreal root of a(s). Since a(-z) = \pm a(z), we may assume that z is in the right half plane. But, since meromorphic functions are open mappings, a small ball around z would then surject to a small neighborhood of 0, and would in particular escape the right half plane. A similar argument applies if z is a non-real root of b(s).

Posted by: David Speyer on January 4, 2023 9:37 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

Sorry, this isn’t quite right. The correct conclusion is that all the roots are on the imaginary axis. I’ll try to write up a correct version once I’m sure there aren’t more errors.

Posted by: David Speyer on January 4, 2023 9:41 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

Okay, here is the proof and then I’ll be quiet for a while. Let f(s)=a(s)/b(s) be in the free slice. Recall that each of a and b is of the form either c(s^2) or s c(s^2) for a polynomial c with positive coefficients, so a(s) and b(s) can’t have real roots (except 0).

Let f(s) = a(s)/b(s) be in the free slice. Suppose for the sake of contradiction that a(s) has a root z which is not purely imaginary. Since a(-s) = \pm a(s), we can assume without loss of generality that z is in the first quadrant. Choose a wedge W(\theta) that contains z in the interior, and choose a small ball B around z lying in W(theta). Then, by the open mapping theorem, f(B) contains an open neighborhood of 0 and, in particular, f(B) is not contained in W(theta). Contradiction.

A similar argument applies to roots of b.

Posted by: David Speyer on January 4, 2023 10:01 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

I believe we can also prove (again with slightly questionable rigor) “Conjecture 1”, that the degree of the associated polynomial to a diagram class containing $n$ rectangles is at most $n$.

Note that a guillotine diagram with $n$ rectangles requires $n-1$ guillotine slices. Therefore such a diagram is constructed using $n-1$ join operations. Hence in the “un-simplified” form of the associated rational expression, there will be $n-1$ additions. For example, the the sixth example for $n = 4$ has $3$ and has an associated expression with three additions:

$x + (x^{-1} + (x + x^{-1})^{-1})^{-1} = x + \frac{1}{\frac{1}{x} + \frac{1}{x + \frac{1}{x}}}.$

We prove that an expression of this form with $k$ additions simplifies to an associated rational expression with numerator and denominator of degree at most $k+1$. We may prove this inductively. The base case is $k = 0$ with associated expression $x/1$.

Now assume that the hypothesis holds for all $n \lt k$. Then for an expression consisting of $k$ additions, we may split the expression at the outermost/topmost addition(thinking in terms of an abstract syntax tree) to obtain an expression of the form $x + r(x)$ or $x^{-1} + r(x)$, where $r(x)$ is the remaining rational expression consisting of $k-1$ additions(in its un-simplified form). Let $p(x)$ and $q(x)$ be the polynomial numerator and denominators obtained after simplification of the remaining expression(by only setting common denominators).

By the induction hypothesis $\max(\deg p(x), \deg q(x)) \leq k$. Then adding back in $x$ or $x^{-1}$ yields $x + \frac{p(x)}{q(x)} = \frac{x q(x) + p(x)}{q(x)} \quad \text{or} \quad x^{-1} + \frac{p(x)}{q(x)} = \frac{x p(x) + q(x)}{x q(x)}.$

It follows that the original expression has numerator and denominator of degree less than or equal to $k+1$. Thus the associated polynomial has degree $\leq k+1$.

Posted by: Jeffery Opoku on January 1, 2023 11:16 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

Note: There is an error where I assumed the other operand is $x$ or $x^{-1}$. The way to actually proceed, I think, is to split the expression at the topmost addition into two subexpressions containing $j-1$ and $k-j$ additions respectively. Then I think one can apply the induction hypothesis and proceed as before.

Posted by: Jeffery Opoku on January 1, 2023 11:31 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

Thanks for this interesting reply! I need to think about it a while longer, but then I’ll have something to say.

Posted by: John Baez on January 4, 2023 12:00 AM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

If one divides an $n$-cube into $n + 1$ similar $n$-orthotopes, what proportions can these $n$-orthotopes have?

### Re: Dividing a Square into Similar Rectangles

For a different approach to this problem, I recommend Lisanne’s 11-part thread on Mathstodon. Don’t worry, it’s not actually long — it’s 11 short posts.

She shows that if you can chop a square into $N$ similar rectangles of proportion $x$, then $x$ is an algebraic number of degree at most $N$.

Here the ‘proportion’ $x$ is the number $0 \lt x \le 1$ which characterizes the similarity class of rectangles. I.e., it’s either the height over the width, or the width over the height, whichever is smaller.

Her analysis is nice because it finds a degree-$N$ polynomial obeyed by $x$ rather explicitly, and it has interesting connections to electrical circuit theory.

Posted by: John Baez on January 6, 2023 8:40 PM | Permalink | Reply to this

### Re: Dividing a Square into Similar Rectangles

A combination of Lisanne’s transformation of the dissection problem to solving Kirchhoff’s equations on a planar graph and Jeffery Opoku’s analysis of guillotine diagrams results in a cute proof of a conjecture in oeis A338781.

The conjecture states that the maximum number of distinct resistances that can be produced from a circuit of $n$ resistors of two different kinds using only series and parallel combinations is divisible by two.

We can see that we need only consider resistances of $x$ and $1/x$ as if we start out with resistances of $u$ and $v$ we can scale them by dividing by $\sqrt{uv}$, thereby making them inverses of each other. The total resistance of a circuit will be scaled by the same amount, thereby leaving the number of distinct resistances the same.

By Lisanne’s construction a series-parallel combination corresponds to a guillotine dissection. The rational function produced by Jeffery Opoku’s method corresponds to the resistance of a series parallel combination of resistances $x$ and $1/x$ as well as the ratio of height to width of the corresponding rectangle.

So we need to prove that the number of distinct rational functions we can make using Jeffery Opoku’s method is even. But this is obvious, the rational function cannot equal one for all $x$, so the resistances of the original circuit and its dual (corresponding to rotating the rectangle 90 degrees or inrverting the rational function) are different, thereby occurring in pairs.

Posted by: Deinst on January 23, 2023 1:04 AM | Permalink | Reply to this

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