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If there is an unit circle and inscribed in it there is a regular n-sided polygon, what is the minimum number of circles with radius $1/2$ to cover the polygon completely?

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What polynomial? Do you mean polygon? –  chubakueno Jan 26 at 0:59
    
yup oops. will edit –  jd123 Jan 26 at 1:01
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The title is still wrong... –  chubakueno Jan 26 at 1:55
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Basically I'm a failure. Hahaha –  jd123 Jan 26 at 2:14
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You might want to look at www2.stetson.edu/~efriedma/packing.html, which gives a whole variety of packing and covering problems, but incidentally not yours. –  user21820 Jan 26 at 3:52

1 Answer 1

Let $m(n)$ be the answer to the question.

For all $n$, you can cover the polygon with $n$ circles located halfway between the center of the polygon and the vertices. Moreover, you can cover the entire unit circle with $7$ circles. So $m(n)\leq n$ and $m(n)\leq 7$.

For $n < 6$, the vertices of the polygon are so far apart that no circle can cover more than one of them. So $m(n)\geq n$, and combined with the previous paragraph, we have $m(n)=n$. With a little finesse, this argument can also be extended to $n=6$: two vertices can be covered by the same circle, but just barely, so each vertex would have to be covered again from another angle anyway.

It is reasonable to conjecture that $m(7)=7$, although I don't have a proof for that. But that's the only remaining task: prove that $6$ circles cannot cover the heptagon... or can they?

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Difficult... -_- What I thought about is if we use the sine rule, than we find that the side length of the heptagon is around .8678 ergo the distance left from the middle of the chord to the opposite end is 1/2<x<1. Pretty awkward to figure out that length. However, I think from there you can say that if the 1/2 radius circle covers two axises, then it cannot also cover the center. –  jd123 Jan 26 at 3:33

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