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:
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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?
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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.