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?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
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?