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You are given two identical pieces of paper of size $a \times \sqrt{2} a$, like the standard DIN A4 paper. Put one paper on top of the other, such that none of the corners is under or above the other paper.

What is the smallest possible intersection area (in terms of $a^2$) that can be achieved?

    To prevent answers:
  • The intersection area would be $0$ if you choose $a = 0$. But I'm asking for the smallest intersection area in terms of $a^2$, which means that the ratio of the intersection area to $a^2$ should be as small as possible.
  • Another way of illegally achieving $0$ intersection area is to put the papers next to each other with some distance in between and declaring that direction vertical. Like this, one paper would be "on top" of the other.
  • Don't fold or cut the paper.
  • ...or basically everything that goes beyond the mathemetical intention of this puzzle.
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Just by eyeballing the possible solutions, it looks like this kind of configuration has the smallest intersection:

enter image description here

That is, H stays on AD, C stays on FG, and the rotation angle is somewhere between 45° and 90°.

Assuming this is the case, the problem reduces to another one, namely:

If you fold one corner (H) of the paper to the opposing long side (AD), how exactly should you fold to minimise the folded area.

That problem can be solved in a lengthy and boring manner (it's problem 61), and the answer is that

OC should be twice the length of DO.

The linked problem also gives a formula for the area, so inputting the solution value,

the area of the fold is $\frac{2 \sqrt{3}}{9}a^2$, so the total area of the intersection is twice that, or $\frac{4 \sqrt{3}}{9}a^2 \approx 0.77a^2$

I really hope a more puzzlingly-like solution exists.

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  • $\begingroup$ I have a solution that is a bit shorter than what you linked, but basically the idea is to show that a certain intersection angle is optimal. About the problem that this could be boring: At some point you need a mathematical formulation in every [optimization] puzzle, because just argumenting doesn't prove that a solution is optimal. $\endgroup$ – A. P. Jan 24 '18 at 11:50

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