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Suppose two convex polygons with $n$ vertices each.

Proposition: There are always $2(n-1)$ ways to intersect them such that these intersections are convex polygons with $3,...,2n$ vertices each.

Is this proposition true?

Is there any theorem related to this?

I found this "fact" drawing something like this:

For $n=4$ For $n=4$

Which other conditions I have to consider for this? Like size for example.

Thanks for your help.

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  • $\begingroup$ Why do you want to reject the edit? I tried to fix some spelling and the title. $\endgroup$
    – null
    Commented Sep 8, 2015 at 22:11

1 Answer 1

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You cannot intersect these two squares to obtain an $8$-gon. The best you can achieve is a $6$-gon (arranging the small square near a corner of the large square).


          TwoSquares
So indeed size must be taken into account.

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  • $\begingroup$ But do you have any clue about how should I consider size? $\endgroup$ Commented Sep 9, 2015 at 4:41
  • $\begingroup$ @IvanS.Guerra: I would assume you have two identical copies of the same convex $n$-gon. Then it may be that your Proposition is true. $\endgroup$ Commented Sep 9, 2015 at 11:49
  • $\begingroup$ Where can I look for more information about this kind of things? $\endgroup$ Commented Sep 11, 2015 at 4:01

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