Geodesics on Regular Polyhedra

As a roleplaying game enthusiast I spend much of my free time interacting with – or boring my friends with talks on – platonic solids. My interest in the geometry of the platonic solids was piqued by a Numberphile video concerning geodesics on platonic solids. Admittedly, my knowledge of geometry is lacking, and so I have committed to writing a blog post on the subject to encourage myself to read the related papers. The papers I will discuss in the following posts are by Fuchs [1], Athreya and Aulicino [2], and Athreya, Aulicino, and Hooper [3].

Over the next few posts I aim to discuss the subject of simple geodesics on platonic solids. This post is intended for a reader with no background in geometry.

The question posed is thus: Given a regular polyhedron, does there exist a simple geodesic starting and ending at the same vertex?

What does this mean in plain English? Consider an ant which lives on a corner on the surface of a polyhedron. Is it possible for the ant to traverse the surface in a straight line, beginning and ending at the same corner, while crossing no other corners?

The five platonic solids are the tetrahedron, cube, octahedron, dodecahedron, and icosahedron. We will consider each platonic solid separately, saving the most interesting case (the dodecahedron) until the end.

The Tetrahedron

To tackle the subject of geodesics on platonic solids, it helps to envision the net of these objects. Consider the net of the tetrahedron, known in gaming groups as the Triforce.

Labelling the vertices ${a}$ , ${b}$ , ${c}$ , and ${d}$ , we tile the plane with this net. Straight line segments on the planar development of the tetrahedron correspond to geodesics on the surface of the tetrahedron. Considering the section of the plane on Figure 2, a line segment on this plane starting at the bottom left corner and ending at some other vertex on the plane is expressed as ${p\pmb{u}+q\pmb{v}}$ , where ${p}$ and ${q}$ are non-negative integers. If both ${p}$ and ${q}$ are even then the line segment ends at ${a}$ ; if ${p}$ is odd and ${q}$ is even then the line segment ends at b; if ${p}$ is even and ${q}$ is odd then the line segment ends at ${c}$ ; and if both ${p}$ and ${q}$ are odd then the line segment ends at ${d}$ .

Figure 2: The planar development of the tetrahedron, source [1]

For a line segment to be simple it is necessary that ${p}$ and ${q}$ are coprime, else a line segment of length ${{kp\pmb{u}}+{kq\pmb{v}}}$ intersects a vertex at ${p\pmb{u}+q\pmb{v}}$ . As a line segment only ends at ${a}$ if both ${p}$ and ${q}$ are even, there exists no simple line segment beginning and ending at ${a}$ .

The Octahedron

Consider the octahedron, which has 6 vertices and 12 edges. We label 3 vertices ${a}$ , ${b}$ , and ${c}$ while we label their opposite vertices ${{a}'}$ , ${b'}$ , and ${c'}$ . The planar development of the octahedron is a bit trickier: we tile the plane as in figure 3, however each vertex on the plane may represent a vertex or it’s opposite vertex on the octahedron e.g. ${a}$ on the plane may represent ${a}$ or ${{a}'}$ on the octahedron. Similarly, each edge on the plane may represent any of 4 edges on the octahedron.

Similar to the case of line segments on the planar development of the tetrahedron, a line segment on the planar development of the octahedron is simple if it has the form ${{p\pmb{u}} + {q\pmb{v}}}$ where ${p}$ and ${q}$ are coprime. A simple geodesic starting at ${a}$ will end at either ${a}$ or ${{a}'}$ if ${p}$ and ${q}$ are congruent modulo 3. Now it must be proven that such a geodesic necessarily ends at ${{a}'}$ . Fuchs proved this by considering a line segment on the plane, and counting the number of edges intersected by this segment which are opposite from a vertex ${a}$ . This number is always odd. For full details of the proof see [1, Theorem 3.1].

Why is the oddness of this number important? Suppose a geodesic starts at ${a}$ and ends at ${a}$ or ${{a}'}$ . If the geodesic intersects an edge opposite from ${a}$ it now falls on a face of the octahedron which has ${{a}'}$ as one of its vertices. If the geodesic ends before intersecting another opposite edge, it must end at ${{a}'}$ . Since the geodesic must intersect an even number of opposite edges to land on a face with ${a}$ as a vertex, it is sufficient to prove that the number of opposite edges intersected by the line segment is odd.

In my next post I will describe the problem and its solution for the cube and the icosahedron.

Bibliography

D. Fuchs, Geodesics on Regular Polyhedra with Endpoints at the Vertices, Arnold Math J. 2, 2016, 201–211
J. S. Athreya, D. Aulicino, A Trajectory from a Vertex to Itself on the Dodecahedron, Amer. Math. Monthly, 126(2), 2019, 161-162
J. S. Athreya, D. Aulicino, W. P. Hooper & with an appendix by A. Rendecker, Platonic Solids and High Genus Covers of Lattice Surfaces, Experimetal Math., 2020

The Tetrahedron

The Octahedron

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