Bobbin' and weavin'

A weaving loom set comes with a square with equally spaced hooks along each of its sides, as well as elastic bands that can be attached to the hooks.

Suppose a particular weaving loom has $N$ hooks on each side, evenly spaced from one corner to another (i.e., there are two hooks on the two corners and $N−2$ hooks between them). Let’s label the hooks along one side $A_1$ through $A_N$, the hooks on the next clockwise side $B_1$ through $B_N$ (with $A_N$ and $B_1$ denoting the same hook), the hooks on the third clockwise side $C_1$ through $C_N$, and the hooks on the final side $D_1$ through $D_N$.

Next, let’s use a whole bunch of elastic bands to connect hooks $A_1$ and $B_1$, $A_2$ and $B_2$, $A_3$ and $B_3$, and so on, up to $A_N$ and $B_N$. When $N = 100$, here’s what the loom looks like:

As $N$ increases, what is the shape of the curve formed by the edges of the bands?

Let's set up the framework, by assuming that the weaving loom, no matter the value of $N$, is the unit square $[0,1] \times [0,1].$ So that for instance $A_i = (0, \frac{N-i}{N-1})$ and $B_i = (\frac{i-1}{N-1}, 0),$ for $i = 1, \dots, N.$ The band connecting $A_i$ to $B_i$ can be represented by the line $$y = \frac{N-i}{N-1} \left(1 - \frac{N-1}{i-1} x\right) = \frac{N-i}{N-1} - \frac{N-i}{i-1} x.$$ Since we are primarily worried about the aggregate curve that is traced out by all of these lines, we really want to have $$f_N(x) = \max_{i = 2, \dots, N} \left\{ \frac{N-i}{N-1} - \frac{N-i}{i-1}x \right\}$$ which will eventually form a continuously differentiable function $f(x) = \lim_{N\to \infty} f_N(x).$

The interval on why the line from $A_i$ to $B_i$ will be maximal is $\left[ \frac{(i-1)(i-2)}{(N-1)^2}, \frac{i(i-1)}{(N-1)^2} \right]$ and the midpoint of this interval is $x_i = \left(\frac{i-1}{N-1}\right)^2.$ If we can find some convex $f$ such that $f_N(x_i) = f(x_i)$ for each $i = 2, \dots, N$ and $$\frac{d}{dx} f(x_i) = -\frac{N-i}{i-1},$$ for each $i = 2, \dots, N,$ then $f(x) = \lim_{N \to \infty} f_N(x)$ for all $x \in [0,1],$ since convex functions are the pointwise supremum of all affine minorants, up to and including their subdifferentials. So in particular, we see that $$-\frac{N-i}{i-1} = -\frac{N-1 - (i-1)}{i-1} = 1 - \frac{1}{\sqrt{x_i}},$$ for each $i = 2, \dots, N.$ Therefore, our best guess would be $$f(x) = f(0) + \int_0^x (1 - \frac{1}{\sqrt{t}}) dt = f(0) + x - 2\sqrt{x}.$$ Since the first band from $A_1$ to $B_1$ is along the $y$-axis, we should define $f(0) = 1,$ so that the weaver's loom always traces out a lower approximation of the convex function $$f(x) = 1 -2\sqrt{x} + x = (\sqrt{x} - 1)^2.$$ In particular, we can verify that $$f_N(x_i) = \frac{N-i}{N-1} - \frac{N-i}{i-1} x_i = \left( \frac{N-i}{i-1} \right)^2 = \left( \frac{i-1}{N-1} - 1\right)^2 = f(x_i),$$ and since $\frac{d}{dx}f(x_i) = -\frac{N-i}{i-1}$ for each $i = 2, \dots, N,$ by design, so we have confirmed that $f(x) \geq f_N(x)$ for all $N$ and $x \in [0,1]$ and the $f_N(x) \uparrow f(x)$ for all $x \in [0,1].$

Let's now go further and quadruple the number of bands placed on the weaving loom. In addition to the band connecting each $A_i$ and $B_i,$ you also place bands connecting $B_1$ and $C_1$, $C_1$ and $D_1$, and $D_1$ and $A_1$. You do this for all the sets of hooks from $1$ through $N,$ so that a total of $4N$ bands have been placed. When $N =100$, here is what the loom looks like:

As N increases, what fraction of the loom’s area lies between the four sets of bands? In other words, what fraction of the square above does the central white region make up?

We can skip to the end with our limiting curve of $f(x) = (\sqrt{x}-1)^2$ from the first portion of the problem, and then by symmetry take the area below the line $y = \frac{1}{2}$ and above $f(x)$ so long as $0 \leq x \leq \frac{1}{2}.$ This should give us one fourth of the answer. In particular, we see that the curve $y=f(x)$ intersects the line $y = \frac{1}{2}$ at $x = \left( 1 \pm \frac{\sqrt{2}}{2} \right)^2 = \frac{3}{2} \pm \sqrt{2}$ where only the negative sign gives a feasible answer less than $x = \frac{1}{2}$. Since we've defined the entire loom to have unit area, the fraction of loom made up by the white area as $N \to \infty$ is given by $$A = 4 \int_{\frac{3}{2}-\sqrt{2}}^\frac{1}{2} \left( \frac{1}{2} - (\sqrt{x} - 1)^2 \right) \,dx = \left[ \frac{16}{3}x^{3/2} -2x^2 -2x \right]_{\frac{3}{2} - \sqrt{2}}^\frac{1}{2} = \frac{8\sqrt{2}-10}{3} = 0.43790...$$