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# Subdividing a square into four similar rectangles
## Problem
Here is an interesting problem [posed by John Carlos Baez ](https://mathstodon.xyz/@Danpiker/109512464477416714): "When chopping a square into 4 similar rectangles, what are the possible aspect ratios?" Below are 11 possible cases—up to rearrangement of the rectangles—[found by Baez and Dan Piker ](https://mathstodon.xyz/@johncarlosbaez/109517010782719784):
![](https://i.imgur.com/VMuauG3.jpg)
Here is the sketch of a proof that the 11 cases above are the only ones.
## Proof
We start by asking: How many of the 4 rectangles share a corner with the main square?
- It cannot be 1, because then the rectangle would have to share all four corners with the main square, which would leave no space for the remaining 3.
- It can be 4: this will lead to case 11.
- It can be 2: this will lead to case 1 and rearrangements of case 2.
- It can be 3: this will lead to all other cases.
Let's start with the simplest case, where each rectangle shares exactly one corner with the square: case 11 is the only possibility, since all other aspect ratios leave space in the middle (a square, possibly with 4 rectangular prongs).
The case where 2 rectangles share corners with the cube leaves more options open. Certainly, it cannot be the case that one rectangle shares three corner with the cube, as this would necessarily force it to share all four corners. We thus have two rectangles sharing two corners each, leaving a rectangle in the middle; without loss of generality, we'll arrange this "sandwich" horizontally. The central rectangle has to be decomposed into two rectangles similar to the boundary ones, and we can categorise the ways in which this is done by asking along which faces the two rectangles touch:
- long face to long face: this is case 1
- long face to short face: this is (a rearrangement of) case 2
- short face to short face: this is (a rearrangement of) case 4
In the case where 3 rectangles share corners with the cube, one rectangle must share 2 corners and hence must be strictly larger than all others: without loss of generality, we'll arrange this largest rectangle at the bottom (this is the red rectangle in cases 1-10), and refer to the remaning area to be covered as the "main sub-rectangle". For the remainder of the proof, we focus on ways to cover the main sub-rectangle using the 3 smaller rectangles (which are constrained to be similar to the large rectangle at the bottom). We write $h$ for the height of the bottom rectangles, so that the main sub-rectangle is $1 \times (1-h)$ (with the constraints $0 \lt h \lt 1$).
One of the 3 smaller rectangles, at least one has two share two corners of the main sub-rectangle, and it can do so either along the long face of the main sub-rectangle or along its short face:
- sharing along the long face of the main sub-rectangle will lead to case 2, case 4 and their rearrangements
- sharing along the short face of the main sub-rectangle will lead to all other cases
If one of the smaller rectangles shares two corners with the long face of the main sub-rectangle, then similarity forces it to be congruent to the large rectangle at the bottom, and we can re-arrange the situation into the sandwich previously discussed (leading to cases 1, 2 and 4). Hence, the only remaining case to be discussed is one where one of the 3 smaller rectangles shares two corners with the short face of the main sub-rectangle. Without loss of generality, we take a largest rectangle doing so (there can be two) and we position it on the left: this is the "blue rectangle" in cases 3 and 5-10.
At this point, there are two remaining combinatorial degrees of freedom: the orientation of the blue rectangle and the orientations of the remaining two rectangles (call them "yellow" and "green", with reference to the figure above). If the blue rectangle is vertically oriented—that is, if the two corners of a long face are those shared with the main sub-rectangle—then we have the following cases, depending on orientation (each horizontal or vertical) and arrangement (horizontal or vertical) of the three rectangles:
- HH orientations and V arrangement: this is case 3
- HH orientations and H arrangement: this is (a rearrangement of) case 9
- VV orientations and V arrangement: impossible by similarity ($\frac{3}{2}h(1-h)-1=0$ has no real roots)
- VV orientations and H arrangement: impossible by similarity ($3h(1-h)-1 = 0$ has no real roots)
- HV orientations: impossible by similarity under both arrangements (share face forces green and yellow rectangles to be squares, but $h \lt 1$)
If the blue rectangle is horizontally oriented—that is, if the two corners of a short face are those shared with the main sub-rectangle—then we have the following cases:
- HH orientations and V arrangement: this is case 7
- HH orientations and H arrangement: this is case 10
- VV orientations and V arrangement: this is case 5
- VV orientations and H arrangement: this is case 8
- HV orientations and V arrangement: this is case 6
- HV orientations and H arrangement: this is case 9
This completes the proof that the original 11 cases are the only possible ones.
## Algorithm to compute the possible aspect ratios (WiP)
The steps in the proof can be used to sketch (part of) an algorithm that computes polynomials whose roots determine the allowed aspect ratios for the main rectangle (generalising the square) and the similar rectangles used to tile it (which might have a different aspect ratio compared to the main rectangle).
Consider a main rectangle of aspect ratio $w \geq 1$; henceforth, aspect ratios are defined to be width:height. We wish to find all possible finite tilings of the main rectangle by a given number $N \geq 1$ of sub-rectangles all of the same aspect ratio $x \geq 1$: for each tiling, we wish to compute a polynomial of which $x$ is a positive real root, subject to a set of polynomial inequalities for $x$ and $w$ (including at least $w \geq 1$ and $x \geq 1$). The original problem is the special case $w=1$ and $N=4$.
Assigning sub-rectangles to the corners of the main rectangle and recursing on a smaller main rectangle gives a skeleton for the algorithm. We have the following cases:
1. There is 1 sub-rectangle sharing all four corners with the main rectangle.
2. There are 2 sub-rectangles sharing two corners each with the main rectangle.
3. There are 3 sub-rectangles sharing corners with the main rectangle, and one of them necessarily shares two corners with it.
4. There are 4 sub-rectangles sharing one corner each with the main rectangle.
Cases 1, 2 and 3 immediately lead to a recursion: **Work in Progress**
Case 4 is significantly more complicated. **Work in Progress**