# Ratio of vertices to edges is always 2:3

I noticed that the ratio of the vertices to edges on shapes with more than one vertice is always 2:3. Is there any equation or mathematical proof that backs this up?

For example, a cube has 8 vertices and 12 edges, and the ratio of vertices to edges is 8:12, which simplifies to 2:3.

• Have you considered a pyramid with a rectangular base?
– Blue
Apr 6, 2020 at 20:24
• This isn't always true - for instance, an icosahedron has 12 vertices and 30 edges. There is a lot of fun math here, depending on what you mean by "shape" - for instance, Euler's formula for polyhedra is a lot of fun to learn about the first time around. This post should probably be edited with some details, though. Apr 6, 2020 at 20:24
• Euler formula $V-E+F=2$ (plus.maths.org/content/eulers-polyhedron-formula) mentionned by @KReiser is fundamental. You are dealing with the case $V/E = 2/3$. There are many other cases... Apr 7, 2020 at 15:18

Assuming your observation is about polyhedra, it is only true if the skeleton of the polyhedron is a $$3$$-regular graph, or alternatively if three edges meet at every vertex. So the octahedron doesn't fit, having $$6$$ vertices and $$12$$ edges.

The $$2:3$$ ratio for polyhedra with three edges meeting at every vertex is an easy consequence of the hand-shaking lemma.

An octoherdron has 6 vertexes and 12 edges.

But there is a rule that relates faces, edges, and vertices. Euler's (other) formula says that all simply connected polyhedra (i.e. no holes) have the same Euler characteristic. That is $$F+V-E = 2$$ For example, a cube has 6 face, 8 vertexes, and 12 edges.

A 1-holed polyhedral torus has $$F+V-E = 0.$$ Additional holes lower the Euler characteristic by 2.

Very easy to find examples and counter-examples: 