Consider a parallelepiped $P$ in $\mathbb{R}^3$ whose vertices are $a_i,\ 1\leq i \leq 8$. If $X$ is an interior point in convex hull of vertices, then define unit vectors $$ b_i = \frac{a_i-X}{|a_i-X|}\in \mathbb{S}^2$$
Then we define a subset $A$ in $\mathbb{S}^2$ :
If $[a_ia_j]$ is an edge in $P$, then $A$ contains a great arc between $ b_i$ and $b_j$ whose length is smaller than $\pi$. $A$ looks a curved cube.
Problem : Then the following is correct ? $${\rm length}\ A\geq 4\pi$$
Example : $(0,0,0), \ (\varepsilon,0,0),\ (0,\varepsilon,0),\ (\varepsilon,\varepsilon,0)$ and $(0,0,1),\ (\varepsilon,0,1),\ (0,\varepsilon,1),\ (\varepsilon,\varepsilon,1)$
Consider a parallelopiped whose vertices are the above 8 points. If we take $X$ as a center, then ${\rm length}\ A$ is close to $4\pi$.
Example : $(0,0,0), \ (1,0,0),\ (0,1,0),\ (1,1,0)$ and $ (0,0,\varepsilon ),\ (1,0,\varepsilon),\ (0,1,\varepsilon ),\ (1,1,\varepsilon)$
Consider a parallelopiped whose vertices are the above $8$ points. If we take $X$ as a center, then ${\rm length}\ A$ is close to $4\pi$.
Equivalent Problem : Problem can be proved by Crofton formula if the following is true :
For any unit vector $U$, there is a two dimensional plane $U^\perp$ containing $X$ s.t. $U$ is orthogonal to the plane $U^\perp$. Then prove that $U^\perp$ meets at least $4$ edges of $P$ in average.
Considere the unit ball centered in the $(0, 0, 0)$ and now imagine a plane $\pi$ intersecting the sphere. This plane intesects the sphere as a circle or a single point and this process can be done continuously. If you intersect this plane creating a circle you can choose 8 points in the border of this circle and considere a curve linking these eight points. Now, since you can reduce the radius of this circle to zero continuously, it means you can reduce the lenght of the curve as well, even lower than $4\pi$ for some curves, as those wich are contained in the circle.