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Here's an horrible drawing that tries to explain what I'm asking:

Okay, so if you're reading or hearing this you can't see images, so I will take the effort of explaining it with words. Consider four $n$-sided polygons arranged so that their centers are the vertices of a square. The square is exactly as large as the diameter of the smallest circle enclosing each polygon. The polygons do not overlap. Polygons are oriented so that each one has either exactly two sides in common or exactly one side and one vertex in common with the other shapes.

Trying this with small numbers gives me $f: 4 \to 0, 6 \to 4, 8 \to 4, 10 \to 8, 12 \to 8.$ This suggests that $$f(2n) = 4 \times \left( \lceil \frac{2n}{4} \rceil - 1 \right), 2n > 4.$$

Can this result be extended for $n \to \infty$? Bonus: What about odd values of $n$?

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I am limiting myself to $2n$-sided shapes because the answer for $2n+1$ sided shapes depends on the exact arrangement. –  badp Oct 14 '10 at 22:10
    
But if you define the arrangement you will find a pattern as well. I suggest you put a flat on the bottom on the bottom row and a flat on the top on the top row and see what you get. The vertices on the left and right will meet that way. Have fun. –  Ross Millikan Oct 14 '10 at 22:16
    
@Ross, I have defined it in the meantime in the alt text for the image, which I'll quote for your convenience. (Blind users, I'm sorry for the repetition.) Consider four $n$-sided polygons arranged so that their centers are the vertices of a square. The square is exactly as large as the diameter of the smallest circle enclosing each polygon. The polygons do not overlap. Polygons are oriented so that each one has either exactly two sides in common or exactly one side and one vertex in common with the other shapes. –  badp Oct 14 '10 at 22:20
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There are two clear cases. When $2n$ is divisible by $4$, each polygon will have a horizontal side and a vertical side touching other polygons. So the number of sides each polygon contributes to the interior is just the number of sides between a horizontal and a vertical side. (Each side of the "interior" polygon belongs to exactly one of the exterior ones.) This is just $(2n-4)/4$, which is then multiplied by $4$ again to give $2n-4$ as the answer.

In the other case, you have the points of tangency being a side and a vertex. We can reduce this to the other case by "stretching" the polygons so the vertices become edges. So the number of sides in this case is $(2n+2)-4=2n-2$. This is your result.

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Yes, and your answer is correct. Following your figure, assume the polygon has a side at the top and bottom. If n is even, so the polygons have a multiple of 4 sides, there will be a pair of sides at the left and right. There will be $n/2-1$ sides between them and four sets of that bounding the central region. If n is odd, there will be points at the left and right. From the point to the flat on top is $(n-1)/2$ sides, again multiplied by four to make the boundary of the central region. This supports your formula, using the ceiling function to combine the cases.

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