This week’s Fiddler is a classic problem.

A weaving loom consists of equally spaced hooks along the x and y axes. A string connects the farthest hook on the x-axis to the nearest hook on the y-axis, and continues back and forth between the axes, always taking up the next available hook. This leads to a picture that looks like this:

As the number of hooks goes to infinity, what does the shape trace out?

Extra credit:If four looms are rotated and superimposed as shown below, what is the area of the white region in the middle?

My solution:

[Show Solution]

Suppose the weaving loom has size $1\times 1$. If a given string starts a fraction $\alpha$ away from the origin along the $y$-axis, so at the point $(0,\alpha)$, it will connect to the point $(1-\alpha,0)$ along the $x$-axis. Therefore, the set of all lines have equations:

\[

y = \alpha-\frac{\alpha}{1-\alpha}x

\]Suppose the limiting shape has equation $y = f(x)$. For a given $x$, $f(x)$ is the largest value of $y$ achievable for one of the lines above at that given $x$. This is illustrated in the diagram below:

In other words, we have:

\[

f(x) = \max_{0\leq\alpha\leq 1} \left( \alpha-\frac{\alpha}{1-\alpha}x \right)

\]It’s clear that the maximum will not occur at a boundary point ($\alpha=0$ or $\alpha=1$), so it must occur at a point where the derivative with respect to $\alpha$ is zero. In other words,

\[

\frac{\mathrm{d}}{\mathrm{d}\alpha}\left( \alpha-\frac{\alpha}{1-\alpha}x \right)

= 1-\frac{x}{(1-\alpha)^2} = 0

\]The optimal choice is $\alpha_\star = 1-\sqrt{x}$, and this leads to

\begin{align}

f(x) &= 1-\sqrt{x}-\frac{1-\sqrt{x}}{\sqrt{x}}x \\

&=1-2\sqrt{x}+x \\

&= (1-\sqrt{x})^2

\end{align}A more symmetric way to express this function is to write it implicitly. If $y=f(x)$, the set of points $(x,y)$ that satisfy the equation above are precisely the points that satisfy the equation

$\displaystyle

\sqrt{x} + \sqrt{y} = 1

$

Let’s dig a little deeper and see if we can learn more about the shape of this curve. Squaring both sides, rearranging, and squaring again, we obtain:

\[

x^2+y^2-2xy-2x-2y+1 = 0

\]This is a quadratic expression in $x$ and $y$, which means it is a conic section! So either a circle, ellipse, hyperbola, or parabola. But which one? The easiest way to check is to examine the eigenvalues of the quadratic part (there will be two eigenvalues). Here is a handy table for reference:

\[

\begin{array}{l|l}

\textbf{eigenvalue property} & \textbf{conic section} \\ \hline

\text{equal eigenvalues} & \text{circle} \\

\text{same sign and nonzero} & \text{ellipse} \\

\text{different sign and nonzero} & \text{hyperbola} \\

\text{one zero eigenvalue} & \text{parabola}

\end{array}

\]

In our case, the quadratic terms are:

\[

x^2+y^2-2xy = \begin{bmatrix}x\\y\end{bmatrix}^\mathsf{T}

\begin{bmatrix}1 & -1 \\ -1 & 1\end{bmatrix}

\begin{bmatrix}x\\y\end{bmatrix}

\]This matrix is singular (determinant is zero), so there is a zero eigenvalue, and the conic section must be a parabola! To make this more evident, we can rearrange the equation a bit more to obtain

\[

\left(\frac{x+y}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\left(\frac{x-y}{\sqrt{2}}\right)^2 + \frac{1}{2\sqrt{2}}

\]If we imagine rotating our curve by $45$ degrees counterclockwise, this would correspond to the transformation $(x,y)\mapsto\left(\tfrac{x-y}{\sqrt{2}}, \tfrac{x+y}{\sqrt{2}} \right)$, so the equation above in rotated coordinates $(\hat x, \hat y)$ is simply

\[

\hat y = \frac{1}{\sqrt{2}}\hat x^2 + \frac{1}{2\sqrt{2}},

\]which is clearly a parabola.

**Tangency.** Based on the animation above, it looks like the line that crosses $(x,f(x))$ is also *tangent to* the curve at that point. To verify this, we can compute the derivative directly and compare it to the slope of the line:

\[

f'(x) = \frac{\mathrm{d}}{\mathrm{d}x}(1-\sqrt{x})^2 = 1-\frac{1}{\sqrt{x}}

\]But the $\alpha_\star$ we found for a point $x$ produces:

\[

\text{(slope of line)} = \frac{-\alpha_\star}{1-\alpha_\star} = \frac{-(1-\sqrt{x})}{1-(1-\sqrt{x})} = 1-\frac{1}{\sqrt{x}},

\]which is a perfect match!

### Extra credit

To find the area of the central region, we can break our shape into pieces as shown below.

The point $x_0$ corresponds to the $x$-value of our curve where $y=\frac{1}{2}$. This is given by:

\[

x_0 = \left(1-\sqrt{\frac{1}{2}}\right)^2 = \frac{3}{2}-\sqrt{2} \approx 0.0858

\]The area of the central part is the total area of the square ($1$) minus four times the shaded area, which leads us to the integral:

\begin{align}

A &= 1-4\left( \frac{1}{2}x_0 + \int_{x_0}^{\frac{1}{2}} (1-\sqrt{x})^2\,\mathrm{d}x \right) \\

&= \frac{2}{3}\left(4\sqrt{2}-5\right) \\[2mm]

&\approx 0.4379

\end{align}I decided to spare you the details of the integration because they’re not particularly interesting. So the white area in the middle occupies approximately $43.8\%$ of the area of the square.