Dodecahedron Space Station

This problem is asking if there is a Hamiltonian path on a dodecahedron. We know that every platonic solid has a Hamiltonian path. Here they are:

Source: Wolfram Mathword

Here is one possible path:

It's important to note that Hamilton path exists iff there's even (except 0) number of vertices that has odd number of edges connected to them.

In this problem all 20 vertices are odd edged. 20 is even, that means path exists.

The dodecahedron may be observed in the vertex graph as three concentric "rings". Starting from any node in either the center "ring" or outer "ring", the "ring" is traversed until the last node of the "ring" is reached, thus completing the "circumference", and then the traveler moves to the next ring. In this manner, all nodes may be visited only once.

A Hamiltonian path exists if every node has an even number of edges or if an even number of nodes has odd number of edges. The dodecahedron satisfies this condition... so it's possible.

Correct me if my intuition here is incorrect but this problem is similar to the Bridges of Königsberg problem and thus Euler's solution to that problem applies to this one. But kudos are due to those who mapped a solution.

This is a variation of the Hamiltonian Path Problem. The original one has no start or finishing points. However, this adds a twist to it as you must determine that one exists with a specific point as starting and finishing point.

Use the coloring principle