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It is obvious that an open annulus in the complex plane: $S = a < |z| < b$ is connected. That is, each pair of point $z_1$ and $z_2$ in it can be joined by a polygonal line.

What is the minimum number of polygonal lines connecting $z_1$ and $z_2$ in S?

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  • $\begingroup$ It seems clear that the maximum should be achieved when the two points are $-a$ and $a$. Then some geometry of tangent lines should get to an answer. Did you try this approach? $\endgroup$
    – Gerry Myerson
    Commented Mar 25, 2011 at 1:33
  • $\begingroup$ Is "minimal number of polygonal lines" = "the minimal number of straight-line segments in a line joining the two points"? $\endgroup$
    – Igor Rivin
    Commented Mar 25, 2011 at 1:38

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You are asking for a regular $n$-polygon for which the circumradius is equal $b$ and the aposthem is smaller than a. Then what is the smallest $n$ such that $$ b \cos(\pi/n) \leq a $$ and hence $$ n \geq \frac{\pi}{\arccos(a/b)} $$

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  • $\begingroup$ Why is the polygon regular? $\endgroup$
    – Igor Rivin
    Commented Mar 25, 2011 at 13:35
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Perhaps I am misinterpreting the question, but if you mean: What is the maximum, over all $z_1$ and $z_2$, of the minimum number of segments in a polygonal line connecting $z_1$ to $z_2$, then it seems there is no bound: A very thin annulus needs an aribitrarily large number of segments to connect diametrically opposed points.

annulus

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  • $\begingroup$ I took the problem to be, bound the number of segments in terms of $a$ and $b$. $\endgroup$
    – Gerry Myerson
    Commented Mar 25, 2011 at 11:31
  • $\begingroup$ @Gerry: Your interpretation makes sense, in which case my figure supports André's answer. $\endgroup$
    – Joseph O'Rourke
    Commented Mar 25, 2011 at 12:22
  • $\begingroup$ @Joseph, I see your figure and Andre's answer as expansions of my comment on the original question, a comment which I hoped OP would pick up on and develop himself into something like Andre's answer. But OP seems to have left the room. $\endgroup$
    – Gerry Myerson
    Commented Mar 25, 2011 at 22:20

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