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Imagine drawing a straight line $l$ through the center of a square piece of paper with area $1$. Now fold the paper along that line.
Q: What is the function for the area covered by the folded piece of paper $A_f(\theta)$, where $\theta$ denotes the orientation of $l$?

This is not a textbook problem, so I don't know how difficult it will be to come up with a solution, but here are my thoughts so far:

The function must have minima when all points of the folded part of the paper are projected on top of all points of the unfolded part, i.e. when $A_f(\theta=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3}{4}\pi,...)=\frac{1}{2}$.
Because of the symmetry, we can restrict our investigation to $\theta \in[0,\frac{\pi}{4}]$.

Here is my strategy:

$1.$ Find the two sets of points, $S_l$ and $S_r$ ("left" and "right" of $l$, respectively), that $l$ divides the square $S$ into, as a function of $\theta$.
$2.$ Reflect $S_l$ around $l$, $S_l \mapsto S_l^*$.
$3.$ Find the area of points that make up the union $S_l^* \cup S_r$.

$1)$ $l=\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)=t\left( \begin{smallmatrix}1\\ \tan\theta\end{smallmatrix}\right)$ so $S_l=\{x,y : |x| \leq \frac{1}{2} \wedge y> x\tan\theta \}$ and $S_r=S \setminus S_l$.

$2)$ I have found the following formula for reflection, $\mathrm{Ref}_l(v) = 2\frac{v\cdot l}{|l|^2}l - v,$ where $v$ is being reflected around $l$. Ideally I would like to replace $v$ with $S_l$, but I've never tried to handle sets this way and don't really know how to proceed from here.

$3)$ Again, I'm not familiar with how this is done (or indeed if it can be done!). It seems to me that there's a long way from the abstract sets to the geometrical figures.


It is very possible that this can be done in a much simpler way; either way, I'd appreciate both alternative methods of solution as well as insights into how my strategy could be executed properly!

Thanks!

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Let $ABCD$ be your square. Let's call $G$ and $H$ the points where $l$ intersects, respectively, the sides $AB$ and $CD$.

We have the two trapezia $GBCH$ and $AGHD$ that both have the same area (half of the original square, so $1/2$).

We now want to reflect $AGDH$ around $l$, let's call $L$ and $K$ the images of $A$ and $B$ after the reflection. The new side $KL$ of the reflected trapezium will intersect $BC$ and $CH$ in one point each ($P$ and $Q$), and $GL$ will intersect $GB$ in one other point ($R$).

The area you need is the area of the half square $+$ the ara of the triangles $RLP$ and $HKQ$.

Now you need just some basic line-line intersections and you are done... if you need I can try to do the calculations, but I'm lazy and it is easy that I do something wrong.

Good luck.

EDIT.

Take it as it is but I think that the total area is: $$ S=\frac{1}{2}+2\tan \theta \left ( \frac{1}{2}-\frac{1}{2}\tan \theta \right )^2 $$

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HINT

You want to reflect lines $ x= \pm 0.5, y= \pm 0,5 $ about $ y = m x , m= \tan \theta .$ Instead of that consider four among the eight rotated lines

$$ \pm x \cos \theta \pm y \sin \theta = 0.5 ; \, \pm x \sin \theta \pm y \cos\theta = 0.5 $$

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