A fly sits on one edge of a cube (not at a vertex) at some point $a$. It is looking for the shortest possible path that leads back to the starting point, passing through all 6 faces of the cube. How does the length of this path depend on the position of the starting point? What is the shortest path?
A developable surface (whether a curved surface, such as a cone or cylinder, or a polyhedron) between two points on the surface, the shortest curve or broken line, when developed into a plane, is also the shortest, i.e., a straight line. (In the case of a polyhedron, the shortest distance must be selected among the different distances corresponding to the various net developments of the polyhedron).

![](https://cdn.mathpix.com/cropped/2024_05_02_312e32f32da8f1a166bfg-1.jpg?height=458&width=706&top_left_y=275&top_left_x=704)

In this case, we need to find a net of a cube that, when developed into a plane, has two points \( P \) and \( P' \) on the boundary of the net that:

1. coincide on the cube after the net is assembled,
2. the straight line connecting them passes through all six faces of the cube in such a way that every point of the line lies on the net.

It is easy to verify that nets of the cube that contain rectangles made of 3 or 4 faces do not meet our conditions, and thus the only net that satisfies our conditions is the one shown in the diagram. In other nets, the connection between \( P \) and \( P' \) can only be achieved with a broken line, which can be shown to be longer on a case-by-case basis.

In the net shown in our diagram, \( P P' \) is parallel and equal to \( A A' \) or \( B B' \), because \( A P \# A' P' \). Since the length of \( A A' \) (or \( B B' \)) is 3 face diagonals, \( P P' \) (where \( P \) can be any point on the edge \( A B \)) is always \( 3a \sqrt{2} \).

The shortest path, therefore, intersects all six edges at a \( 45^\circ \) angle and forms a hexagon on the cube, whose sides are alternately parallel to the sides of the equilateral triangles \( A C H \) and \( B E G \), meaning the path lies in a plane parallel to the plane of \( \triangle A C H \) (and \( \triangle B E G \)). In other words: the shortest path is always a hexagonal section passing through the point \( P \) and perpendicular to the body diagonal \( D F \), and the perimeter of this hexagon is, according to the above, always \( 3a \sqrt{2} \), regardless of the position of \( P \).