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When I looked at an image of a regular octahedron, I found that its surface was composed of triangles. Then I looked at a regular icosahedron and this was the same case!

Then I saw a regular dodecahedron, and then it was different - it was seemingly made up of pentagons! Why is this so? Why is a regular polyhedra with say $V$ vertices, $F$ faces and $E$ edges visibly made up of these two-dimensional triangles or $n$-gons? I'd be grateful if anyone out there can answer my query. Thanks!

This image will clarify what I'm talking about - https://www.google.com/search?site=&tbm=isch&source=hp&biw=1280&bih=629&q=polyhedra+regular&oq=polyhedra+regular&gs_l=img.3...5507566.5510775.0.5510949.17.13.0.1.1.0.479.1175.2-3j0j1.4.0....0...1ac.1.64.img..12.5.1176.qrusVJCv9aI#imgrc=puHp4xeA9XGs8M%3A

It shows the different polyhedra that there are, and how they can seem to be composed of these two-dimensional triangles or $n$-gons.

soupynoodles
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mathster
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1 Answers1

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Let $n$ be the number of sides of the regular polygons forming the faces of our regular polyhedron ($n\ge3$), and let $k$ be the number of faces meeting at each vertex ($k\ge3$). The total number of polygon sides must be twice the number of edges, because every edge is formed by two polygon sides sticking together, so we have: $$ nF=2E,\quad\hbox{that is:}\quad E={nF\over2}. $$ In a similar way, the total number of polygon vertices must be $k$ times the number or vertices in the polyhedron, because $k$ polygons meet at every polyhedron vertex. Hence: $$ nF=kV,\quad\hbox{that is:}\quad V={nF\over k}. $$ Now combine these two equations with Euler's formula $F+V-E=2$ to get: $$ \tag{1} F+{nF\over k}-{nF\over2}=2,\quad\hbox{that is:}\quad F={4k\over 2k-(k-2)n}. $$ From this you can see that only certain values of $n$ and $k$ are allowed. Suppose for instance we want to find which polyhedra can exist with $k=3$. If you plug this into $(1)$ you get $$ F={12\over 6-n} $$ and so $n$ can only be $3$, $4$ or $5$ for the formula to hold, because $F$ must be a positive integer number. You can continue by yourself: take $k=4,\ 5, \ldots$ and see what happens.

Intelligenti pauca
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