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I would like to understand the deep meaning of a solution which arises from an optimization problem discussed in a paper of mine since it can be simply stated as $\frac{\phi^5}{2}$, where $\phi := \frac{1+\sqrt{5}}{2}$ is the well-known golden ratio.
The problem asks to join all the vertices of the cube $[0,1]^3 \subset \mathbb{R}^3$ (i.e., working in the Euclidean space) by using a polygonal chain consisting of the minimum number of line segments. Then, under the stated constraint, we are going to minimize the volume of the axis-aligned bounding box (AABB) which contains the above-mentioned polygonal chain with minimum link-length.
In order to solve the problem, we have proven that, for each integer $k > 1$, the minimum-link polygonal chains joining all the points of the set $\{0,1\}^k$ have (exactly) $3 \cdot 2^{k-2}$ edges (see Theorems 2.2 and 2.3), and thus our optimal polygonal chains consist of $6$ line segments.

I initially thought that the original question was sufficiently articulated to prevent the existence of a very compact solution.
Surprisingly, in recent years, I was finally able to determine that the polygonal chain $(0,1,0)-(0,0,0)-\left(\frac{1+\phi}{2},0,\frac{1+\phi}{2}\right)-\left(\frac{1}{2},1+\phi,\frac{1}{2} \right)-\left(\frac{1-\phi}{2},0,\frac{1+\phi}{2} \right)-(1,0,0)-(1,1,0)$ minimizes the volume of the AABB for the given problem (for a partial solution, see General uncrossing covering paths inside the Axis-Aligned Bounding Box, pp. 164-165).
Consequently, an optimal AABB is $\left[\frac{1-\phi}{2}, \frac{\phi+1}{2} \right]\times \left[0, 1+\phi \right]\times \left[0, \frac{1+\phi}{2} \right]$, which implies a volume of $\left(\frac{1+\phi}{2}-\frac{1-\phi}{2} \right) \cdot (1+\phi) \cdot \left(\frac{1+\phi}{2} \right)$.
Hence, the minimum volume of the AABB is $\frac{\phi \cdot (1+\phi)^2}{2}$, which can be compactly rewritten as $\frac{\phi^5}{2}$ (i.e., $\frac{11+5 \cdot \sqrt{5}}{4} \approx 5.54508497$).

My conclusion is that this outcome depends on the definition of $\phi$ as the positive root of the quadratic equation $1-x = \frac{1}{x}$ so that we can rearrange a product of three factors where $\phi$ appears trice (inside all the terms describing the lengths of the sides of the AABB) as we go up/down in the hyperoperators ladder by using the mentioned property $1-\phi = \frac{1}{\phi}$. Here, accidentally, we get a simplification by going upstairs on the mentioned ladder.
If so (and this is just an attempt to think about the effects of a property of $\phi$ rather than a deep reasoning on its meaning) the remaining subquestion would ask us to explain why $\phi$ appears as the key value for the optimal polygonal chain of $6$ links that joins all the vertices of a unit cube (and I can anticipate that we get a similar outcome also by considering circuits instead of trails).

Very curious.

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    $\begingroup$ P.S. I've seen that the constant $\phi^5$ is also the "critical activity" of the Hard Hexagon Model (see the OEIS sequence A244593). $\endgroup$ Commented Jul 7 at 11:39
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    $\begingroup$ You could consider posting an answer to the MO question, Golden ratio in contemporary mathematics. $\endgroup$ Commented Jul 7 at 12:05
  • $\begingroup$ Thank you very much, done! $\endgroup$ Commented Jul 7 at 12:14
  • $\begingroup$ The formula $1-\phi=\frac{1}{\phi}$ is the $n=0$ case of the generalized golden ratio relations $\phi_n-\phi_{n+1}=\frac{1}{\phi_{n+1}}$ where $\phi_{-1}=0, \phi_{n+1}=F_+(\phi_n)$ and the golden ratio map $F_+(x)=\frac{x+\sqrt{x^2+4}}{2}=x+\cfrac{1}{x+\cfrac{1}{x+\ldots}}$, see mathoverflow.net/q/474134. Maybe you can generalized your problem. $\endgroup$
    – CHUAKS
    Commented Jul 7 at 13:44
  • $\begingroup$ Thank you @CHUAKS. Maybe it would be interesting to study the same problem in higher dimensions too (e.g., for the planar case, the solution is a trivial spanning path and thus the volume of the AABB is $1 = \phi_{-1} + 1$). $\endgroup$ Commented Jul 8 at 10:47

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