Find minimal lengh of a specific curve in $\mathbb{R}^3$ so a few months ago i competing in a math competion at university level,this competion has 3 types problem ,Analysis,Algebra and something that they called Creative type problem which i think they mean this problems can be solved using simple tools but they need creatively solution.
basically i know the solution of the problem which realsed after contest in other language and i will post it here if in case any one want to read it.
but my main question is

does any body know other solution for this problem,or any book which included this question in it, or even any paper?

also if u know any other solution rather than the solution i posted i will be really greatfull if u share it with me thanks.

Problem:Suppose $M$ is cube , with lenth of $1$ for each edge,and suppose $\gamma$ is a closed curve on the cube sides and touches all the sides(if curve touches an edge or a vertex it is also touches the sides which means if it is touch an edge it actually touchs $2$ sides at a time and if it touch a vertex it actually touch $3$ sides at a time). what is the minimum lengh of the curve?

it is so obvious the minimum lenth is $3 \sqrt2 $ but we need to proof it.
(i translated the solution and i know there is some mistakes in my translation so feel free to edit it)
Solution: without loss of genarality consider $[0,1]^3$ , and suppose $\gamma(t)=(\gamma_1(t),\gamma_2(t),\gamma_3(t)),0\le t \le 1$ for the desired curve.also we can approximately find the lenght using piecewise differentiable curve or polygon.so without loss of genarality we could assume $\gamma$ is a piecewise differentiable curve (or a polygon).
consider the projection of the curve on the $x$ axis we can write $\int _0^1 |\gamma^{'}_1(t)|dt \ge 2$. because the projection is continuous closed curve which cross from $0$ and $1$ so the minimum lenth is $2$.we get the same reuslt for $\gamma_2$ and $\gamma_3$.
also we know that curve touch the side of cube so for the differentiable points of $\gamma$ at least one of the $\gamma_i$ are $0$ using this fact and $\sqrt{a^2+b^2}\ge {{a+b} \over \sqrt{2}}$ we can write:
$$ \sqrt{ \gamma^{'}_1(t)^2+\gamma^{'}_2(t)^2+\gamma^{'}_3(t)^2 }  \ge {{|\gamma^{'}_1(t)}|+{|\gamma^{'}_2(t)}|+{|\gamma^{'}_3(t)}| \over 
 \sqrt 2}$$
taking integrate form both sides one can write:
$$\int _0^1 |\gamma^{'}(t)|dt \ge {1 \over \sqrt 2} \sum _{i=1}^3 \int _0^1 |\gamma^{'}_i(t)|dt \ge 3 \sqrt 2$$
so this is the minimal lengh and it is a triangle which include 3 diameter of cube with lenght $\sqrt 2$.
 A: i) Assume that $\gamma$ is a simple closed curve touching six faces.
Consider the case : For $\varepsilon$, there is a sequence
$\gamma_n$ s.t. (1) $\gamma_n$ is in $\varepsilon$-tubular
neighborhood of $\gamma$, (2) a simple closed curve touching six
faces, (3) $$ {\rm length}\ \gamma'=\lim_n\ {\rm length}\ \gamma_n <
{\rm length}\ \gamma$$
where $\gamma'$ is a pointwise limit of $\gamma_n$, and (4)
$\gamma'$ is locally minimizing except finite points.
Then each projections of $ \gamma'$ onto coordinate planes has
length at
least $2$, by an assumption of $\gamma$.
ii) Assume that $\gamma'$ is in one face $F$ of cube. Note that it
is impossible, since
$\gamma$ can not go into opposite face of $F$.
Hence assume that $\gamma'$ is a closed curve of length
$2\sqrt{5}$.
That is, it can be explained by shortest curve passing through
$(0,0),\ (2,1)$ in a rectangle $$ R={\rm conv}\ \{ (0,0),\ (2,0),\
(0,1),\ (2,1)\}
$$ in
$xy$-plane. Here let $R=I\ \bigcup \ I+(1,0)$.
Here we have a claim that $\gamma'$ is one we will find.
iii) Here note that projections of $\gamma'$ in $I,\ I+(1,0)$ onto
$x$-axis must have length at least 2. And projections of $\gamma'$
in $R$ onto $y$-axis must have length at least 2. This completes the proof.
