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In the Euler formula, for counting the number of faces, we count the regions bounded by edges, including the outer, infinitely-large region, so in the graph $K_1$ there is only one face which is outer (if I have understood it properly); similarly for $K_3$, there are two faces one bounded by edges and the other unbounded regions. In both these cases Euler formula is satisfied.

But when considering the idea of counting faces in Platonic solids, I just don't understand how are we accounting for the outer-face. I guess my instructor explained it with something like mapping into a sphere and back and he says this implies that we can choose to make any face the outer but I didn't understand this approach.

I can see the Euler formula working for (which it should) for the regular polyhedra, but I just can't see how are we taking in account the outer face.

Could anybody help me in understanding these outer-face intuition for these Platonic solids?

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Think of what happens when you flatten a polyhedron into its corresponding graph. You stretch one of the faces so that most of the other vertices are within the vertices of the face you just stretched. Then you flatten down your polyhedron... – J. M. Sep 24 '11 at 4:04
Another way of putting it: treat the polyhedron with one face removed as being made of (extremely stretchable) rubber. You try to turn it inside out and squish it so it flattens. The outer face corresponds to the removed face that allowed you to reach into the rubber polyhedron to stretch it inside out... – J. M. Sep 24 '11 at 4:07
@J. M:Sorry,but I still don't understand how does this account for the outer face (infinitely large region)?! – Quixotic Sep 24 '11 at 8:08
@J.M.: The "infinitely large" idea is that when the graph is put in the Cartesian plane it partitions it into regions, and one of those regions has infinite area. – anon Sep 24 '11 at 8:44
@anon:Hey thanks I got it from your comment! :) – Quixotic Sep 24 '11 at 12:46
up vote 3 down vote accepted

First, your typical looking cube graph or whatever it's called. Then, you put the graph on a sphere (use your imagination for the other side), with the gold dot representing the "point at infinity" under stereographic projection. Oh, but let's put it in the plane - where'd the back face go I wonder?


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Here's a graphical elaboration of what I was talking about in the comments, using the dodecahedron as an example:

expanding dodecahedron

(I might consider doing animations for the other polyhedra if asked nicely...)

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+1,Thanks for the animation,I understood it now.I really don't remember did I ever asked something from you in a way that is not nice? :) – Quixotic Sep 24 '11 at 12:48
I don't think there has been a time where you didn't ask nicely... ;) – J. M. Sep 24 '11 at 12:50
I think I should accept anon's answer as I understood this thing from his comment,however now I can see your point too :) – Quixotic Sep 24 '11 at 12:51
Thanks! I am grateful ;) – Quixotic Sep 24 '11 at 12:52

It is not just Platonic solids, but the whole plane. If you take the plane $\mathbb{R}^2$ and add a point at infinity you have a 2-sphere like the surface of the earth. The "far boundary" all connects to the "point at infinity". I think this is a difficult concept-you can try Wikipedia and see if it helps.

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Sorry but I don't think I am getting the idea at-least as of now :( – Quixotic Sep 24 '11 at 8:09

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