Some background to this problem: Some baths can be remade into showers by pulling on two glass panels. Since my glass panels have high friction with the floor, I would like to sweep as small area as possible while extending them whenever I want to take a shower. I think it can be used also while trying to asses minimal space required for robotic arm to extend properly.

More formally if we look at the situation from above we have two segments of length $1$. First one is connected to the origin and the second one is connected to the second endpoint of the first one. At the start they are folded such that first segment goes from $[0,0]$ to $[0,1]$ and the second one goes from $[0,1]$ to $[0,0]$. Then after extending first segment should go from $[0,0]$ to $[1,0]$ and second segment from $[1,0]$ to $[2,0]$. Extension is any continuous movement of the two segments.

Swept area is total area which have the two segments covered and we count each crossed point exactly once, so in my example if we had kept point $D$ on $x$ axis and just pulled it to $[2,0]$ the total swept area would be the black region.

Now comes the question: What is the minimal area one can sweep while extending the segments?

Also how big is swept area from my example? Rules for my example are that point $D$ (endpoint of second segment) always stays on the $x$ axis and we just pull it along creating many equilateral triangles in the process. There is also a more general version, where we have $1000$ segments where ones with odd number go from $[0,0]$ to $[0,1]$ and ones with even number go the other way around. Notice that my example works also for arbitrary number of segments, since we can just unravel the segments in pairs and pushing those already unraveled along the $x$ axis, giving the same area overall.

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    $\begingroup$ My guess is that you can reduce the area by first extending the second segment by about 60 degrees, then keeping it pointing roughly towards (2, 0) as you extend the first segment. You can't reduce the leftmost black quarter-circle, but this could make the remaining "tail" thinner. $\endgroup$ – Rahul Mar 9 at 6:00

Thanks for this fun question. Partial solution: I worked out the area swept out in your example--I think the answer is $5\pi/16$ (though I may have easily made a mistake).

Let $\psi\in[0,\pi/2]$ denote the angle between the line segment $AC$ and the $x$ axis. The area we seek is the integral of some function $f:[0,2]\to\mathbb{R}$ (illustrated in your figure). I worked out a precise form for $f$ as follows: feach angle $\psi\in[0,\pi/2]$, defines a unique hat function (the line segment $ACD$) given by $$ h_\psi(x)=\begin{cases} \tan(\psi)x,&x\in[0,\cos(\psi)],\\ -\tan(\psi)x+2\sin(\psi),&x\in[\cos(\psi),2\cos(\psi)], \end{cases} $$ and $0$ outside the interval $[0,2\cos(\psi)]$. The function $f$ is then given by $$ f(x)=\max_{\psi\in[0,\pi/2]}\big\{h_\psi(x)\big\}. $$ For any $x\geq1$, we have $\cos(\psi)\leq{x}$, and hence the maximum of $h_\psi(x)$ with respect to $\psi$ can be computed by taking the derivative of $-\tan(\psi)x+2\sin(\psi)$ w.r.t. $\psi$ and setting equal to zero, which yields $\hat{\psi}=\cos^{-1}\big((x/2)^{1/3}\big)$, and thus $$ f(x)=-\tan\big[\cos^{-1}\big((\tfrac{x}{2})^{1/3})\big)\big]\cdot x+2\sin\big[\cos^{-1}\big((\tfrac{x}{2})^{1/3}\big)\big] $$ for $x\geq1$. In fact, this holds for $x\geq1/\sqrt{2}$ (working out the formal details soon). For $x\leq1/\sqrt{2}$, $f$ is just the arc of a circle: $$ f(x)=\sqrt{1-x^2}\text{ for }x\leq1/\sqrt{2}. $$ We can then integrate this closed form expression for $x$ over $[0,2]$. We have that $$ \int_{0}^{1/\sqrt{2}}\sqrt{1-x^2}dx=\frac{\pi+2}{8} $$ and $$ \int_{1/\sqrt{2}}^2-\tan\big[\cos^{-1}\big((\tfrac{x}{2})^{1/3})\big)\big]\cdot x+2\sin\big[\cos^{-1}\big((\tfrac{x}{2})^{1/3}\big)\big]dx=\frac{3\pi-4}{16} $$ and thus $$ \int_0^2f(x)dx=\frac{\pi+2}{8}+\frac{3\pi-4}{16}=\frac{5\pi}{16}. $$ (Integrals were evaluated with Mathematica). I will fill in the missing details soon.

Here's a plot of the function $f$ as I have derived it (the vertical line indicates the separating point $x=1/\sqrt{2}$:

the curve f

Note on the general problem: this can be posed as a problem in the calculus of variations, which can be used to prove that you have indeed found a minimal area.

Sliding hinge


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