Background
I've recently been devising a puzzle that incorporates elements from Egyptian fractions, magic squares, and tilings. The objective of the puzzle is to tessellate a square with sides of length $1$ with tiles that have the surface area of unit fractions. A sum of distinct unit fractions is called an Egyptian fraction.
Let's call an 'Egyptian unity sum set' (EUSS) a set of distinct positive integers $\{ a_{1}, a_{2}, \dots , a_{n} \}$ of size $n$ such that their Egyptian fraction sum to $1$. So we have $$ \frac{1}{c_{1}} + \frac{1}{c_{2}} + \dots + \frac{1}{c_{n}} = 1. \tag{1}\label{1} $$
For instance, $\{2,3,6\}$ is an EUSS for $n=3$.
Moreover, we say that an Egyptian unity sum set is composite if $c_{1}, \dots , c_{n}$ are all composite numbers. An example of a composite EUSS is $\{4, 6, 8, 9, 10, 12, 15, 18, 24\}$. In this case, $n=9$. There are no such sets for $n<9$. But for every $n \geq 9$, there is at least one. More information can be found in this question.
The Puzzle
We will be tiling unit squares with rectangles that each have the surface area corresponding to the reciprocal of the elements of the composite Egyptian sum sets. The sides of the rectangles are also of unit fraction length. The lengths must correspond to a factorization of the multiplicative inverse of the numbers in the composite Egyptian sum sets. That is, for a composite EUSS $\{c_{1}, \dots , c_{n}\}$ we have:
\begin{align*}1 &= \ \ \ \ \ \ \ \ \ \frac{1}{c_{1}} \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \frac{1}{c_{2}} \ \ \ \ \ \ \ \ + \ \ \ \ \dots \ \ \ + \frac{1}{c_{n}} \newline &= \left(\frac{1}{a_{1}} \times \frac{1}{b_{1}} \right) + \left(\frac{1}{a_{2}} \times \frac{1}{b_{2}} \right) + \dots + \left(\frac{1}{a_{n}} \times \frac{1}{b_{n}} \right) \tag{2}\label{2} \end{align*} Equation \eqref{2} describes some composite EUSS factorization. The $n$ rectangles thus have dimensions $\left(\frac{1}{a_{1}} \times \frac{1}{b_{1}}\right), \dots , \left(\frac{1}{a_{n}} \times \frac{1}{b_{n}} \right)$. Here, we require that $a_{i} \neq 1$ and $a_{i} \neq c_{i}$ for all $i$. The same holds for all $b_{i}$. If there are multiple possible factorizations of a number (like $12 = 3 \times 4 = 2 \times 6 $), then one may use tiles with unit fraction lenghts that correspond to the factorization of your own choosing.
Finally, we require that all the rectangles are used exactly once to tile the unit square. So no double tiles or missing tiles are allowed. A tiling with all such rectangles used only once is called a magic Egyptian tiling (MET).
Two non-examples
To clarify the puzzle, it might prove instructive to depict some (none-)examples. Here, we have $n=9$. We will try to obtain an MET for the following composite EUSS factorization: \begin{align*} 1 &= \ \ \ \ \ \frac{1}{4} \ \ \ \ \ + \ \ \ \ \ \frac{1}{6} \ \ \ \ \ + \ \ \ \ \ \ \frac{1}{8} \ \ \ \ \ + \ \ \ \ \ \ \frac{1}{9} \ \ \ \ \ + \ \ \ \ \frac{1}{10} \ \ \ \ \ + \ \ \ \ \frac{1}{12} \ \ \ \ \ + \ \ \ \ \frac{1}{15} \ \ \ \ \ + \ \ \ \ \frac{1}{18} \ \ \ \ \ + \ \ \ \ \frac{1}{24} \newline &= \left( \frac{1}{2} \times \frac{1}{2} \right) + \left( \frac{1}{2} \times \frac{1}{3} \right) + \left( \frac{1}{2} \times \frac{1}{4} \right) + \left( \frac{1}{3} \times \frac{1}{3} \right) + \left( \frac{1}{2} \times \frac{1}{5} \right) + \left( \frac{1}{2} \times \frac{1}{6} \right) + \left( \frac{1}{3} \times \frac{1}{5} \right) + \left(\frac{1}{3} \times \frac{1}{6} \right) + \left(\frac{1}{2} \times \frac{1}{12} \right). \tag{3}\label{3} \end{align*}
Below, I show two non-examples of METs that I've made for this composite EUSS factorization. The size of each of the tiles is annotated in green:
In the image on the left, we see that it is almost an MET, but not quite. The remaining blue tile of size $\frac{1}{15}$ does not fit in the remaining red space of the unit square. Moreover, the lengths of the sides of the red rectangle are not both unit fractions.
In the image on the right, all rectangles do tile the unit square. However, there's trouble in paradise: some of the tiles are used twice. The tiles of size $\frac{1}{12}$ and $\frac{1}{18}$ are both employed twice. Moreover, the tiles of size $\frac{1}{10}$ and $\frac{1}{15}$ are missing. This is not allowed.
Two actual examples
In the following MSE question, which can be considered as a predecessor to this one, user joriki provides an answer to the question:
Does an MET exist for the composite EUSS factorization described in equation \eqref{3} ?
It turns out, my efforts were in vain. With the code he devised - which encompasses a brute-force algorithm that tries all possible configurations of tiles for a given composite EUSS factorization - he finds there are no METs for $n=9$.
Fortunately, there are METs when $n=10$. Below, I depict two examples:
The MET on the left is based on the EUSS $\{4,6,8,9,10,12,14,18,28,840 \} $. Note that it splits the unit square in two, along the horizontal axis. The MET on the right is based on the EUSS $\{ 4,6,8,9,10,12,15,20,24,180 \}$. It does not split the unit square in two: there is a rectangle of size $\frac{1}{180}$ in the middle.
Questions
It turns out that for $n=10$, there are $46$ composite EUSS, and $11$ of them allow for at least one MET.
Define the ratio $$r(n) := \frac{ | \{ \text{All composite EUSS of length } n \text{ that allow for at least one MET} \} | }{ | \{ \text{All composite EUSS of length } n \}| } .$$
So we have $r(10) = \frac{11}{46}$. I have the following questions:
- What happens to $r(n)$ when $n \to \infty$ ? Does it converge to some number, and if so, what is this number?
- What can we say about the asymptotic growth rate of $r(n)$ as $n$ tends towards infinity?
