Connecting points in an annulus with horizontal and vertical segments

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.34

I guess the answer is to do with Eg 1.13:

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For $D[0,1]$ or $D[0,3]$, maximum is 2.

For $D \cdot [0,1]$ or $D \cdot [0,3]$, maximum is 3, achieved when the points are on a line passing through 0. Edit: Upon reflection, I think maximum is 5

For $D[0,3] \setminus D[0,2]$, maximum is I guess 5, achieved when, but not only when, the points are on a line passing through 0 and close to $D[0,3]$?

If right, then what's the justification please?

If wrong, why and how else can I approach this please?

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Asked here but there are no posted answers: Find the maximum number of horizontal and vertical segments in $G$ needed to connect two points of $G$.

Related:

Prove that $A_{r,s}$ $=[z\in \mathbb C : r<|z-z_0|<s ]$ is path connected. An annulus in $\mathbb R^2$ is path connected

Answer 1

(Following up on a previous comment, posting as an answer for the sake of the pic.)

It is possible to draw a $12$-side polygon with only h/v sides contained within the given annulus, and each of the two points can reach it with (at most) one segment. Therefore an upper bound to the answer is $\,8\,$, but this alone doesn't prove that it's the lowest upper bound i.e. the maximum.

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[ EDIT ]   The minimum is in fact $\,5\,$, as shown in Christian Blatter's answer.

A related, more general question, would be what is the required number of segments to connect any two points in an annulus by arbitrary line segments, without restricting them to be horizontal or vertical, only. A similar argument could show that the magic number is $\,\lfloor n/2 \rfloor +3\,$ where $\,n\,$ is the minimum integer such that a regular $n$-gon can be strictly entirely inside the annulus.

Since the ratio between of the radii of the inscribed vs. circumscribed circle to a regular $n$-gon is $\,r / R = \cos \pi/n\,$, it follows that for $\,n=4\,$ an annulus with $\,1/2 \le r/R \lt 1 / \sqrt{2}\,$ would require $\,\lfloor 4/2\rfloor + 3=5\,$ segments (which is the answer here because $2/3 \in \left[1/2, 1/\sqrt{2}\right)\,$ indeed), for $\,n=5\,$ an annulus with $\,1 / \sqrt{2} \le r/R \lt (1+\sqrt{5})/4\,$ would also require $\,5\,$ segments (but those would no longer be just horizontal and vertical), for $\,n=6\,$ an annulus with $\,(1+\sqrt{5})/4 \le r/R \lt \sqrt{3} / 2\,$ would require $\,6\,$ segments etc.

Answer 2

Exercise 1.34 is clearly meant to be a little brain teaser. It has nothing to do with complex analysis or topology.

Choose an $h\in\>\bigl]2,{3\over\sqrt{2}}\bigr[\>$, e.g., $h:=2.06$, and draw the four lines $x=\pm h$, $y=\pm h$. Their four points of intersection are lying in the interior of the annulus $\Omega$. Every point in $\Omega$ can be joined within $\Omega$ to one of these lines by a horizontal or vertical segment. It follows that any two points in $\Omega$ can be joined with a chain of $\leq5$ horizontal or vertical segments $\sigma_i\subset\Omega$.