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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.

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    $\begingroup$ Have you considered a pyramid with a rectangular base? $\endgroup$
    – Blue
    Apr 6, 2020 at 20:24
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    $\begingroup$ 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. $\endgroup$
    – KReiser
    Apr 6, 2020 at 20:24
  • $\begingroup$ 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... $\endgroup$
    – Jean Marie
    Apr 7, 2020 at 15:18

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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.

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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.

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Very easy to find examples and counter-examples:

enter image description here

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