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?
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To prevent lateral-thinking 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.