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The famous statement that

Only five convex regular polyhedra exist.

is usually proven as follows:

Let $P$ be a convex regular polyhedron with

  • V vertices,
  • E edges and
  • F faces.

Moreoever, let

  • n be the number of edges of each face and
  • c be the number of edges which meet at a vertex.

Then $F=\frac{2E}{n}$ and $V=\frac{2E}{c}$ and $c\geq 3, n\geq 3$.

Substituting this into Euler's polyhedron formula $$ V-E+F=2, $$

and doing some easy calculations, one gets that only

$$ (V,E,F)\in\{(4,6,4),(8,12,6),(20,30,12),(6,12,8),(12,30,20)\} $$ are possible combinations.

So far, so good.

But where do we need that $P$ is convex and regular?

(I think the regularity is at least used for $n\geq 3, c\geq 3$, isn't it?)

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    $\begingroup$ Regularity at least implies that $n$ resp. $c$ are constants. That is, their value does not depend on the face resp. vertex. The non-convex genus zero polyhedra that have, say squares as their faces, have varying $c$. At least the ones I can think of right away :-) $\endgroup$ – Jyrki Lahtonen Jul 20 '17 at 19:13
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As already said in the comments, regularity means being composed of equal faces, thus enabling to connect the numbers $V$, $E$ and $F$ by some algebraic relations. This is the left part of Euler's identity $$V-E+F=2$$ Now, convexivity is, in fact, the right-hand side.

In general, for a surface $S$, the formula reads $$V-E+F=\chi(S)$$ where $\chi$ is the Euler characteristic (defined by the equation above or, alternatively, by the alternating sum of dimensions of homology groups).

If a polyhedron is convex, it can be proven that it's boundary is homeomorphic (topologically equivalent) to a sphere $\mathbb{S}^2$, and $\chi(\mathbb{S}^2)=2$, providing the right part of Euler's equation.

So, convex is just a simplification; the classification really works for all polyhedra homeomorphic to a ball. For some other topology, a different classification may arise.

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  • $\begingroup$ That is, if we would not assume convexity, we would have another right-hand side in the Euler formula and would get some other classification different from the five Platonic solids? $\endgroup$ – Rhjg Jul 20 '17 at 20:11
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    $\begingroup$ @Rhjg Exactly! For example, there is an infinite number of toroidal polyhedra, in which case $\chi=0$. $\endgroup$ – lisyarus Jul 20 '17 at 20:16
  • $\begingroup$ Thanks for your explanation, it helped me. $\endgroup$ – Rhjg Jul 20 '17 at 20:16
  • $\begingroup$ compact but clear answer, thanks for providing it. $\endgroup$ – G Cab Jul 20 '17 at 20:50
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The regularity is used to reduce consideration to a single face or vertex because they are all similar. Convexity comes from Euler's formula used with regularity implicitly. That is, since every vertex is similar to every other vertex, the solid must be convex at every vertex if it is convex at one vertex, otherwise if is not convex at a vertex then all vertices are not convex and Euler's formula must be modified.

You can read the Wikipedia article Euler characteristic for more details about the role of convexity and it also links to Wikipedia article "Proofs and refutations" which is very relevant.

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  • $\begingroup$ I am not sure if I do understand your comment about convexity. Where in the sketched proof is convexity needed? Sorry if you already said it. $\endgroup$ – Rhjg Jul 20 '17 at 19:49
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    $\begingroup$ @Rhjg The four non-convex regular polyhedra do not satisfy Euler's formula. For example the great dodecahedron has $12$ vertices, $30$ edges and $12$ faces $\endgroup$ – Henry Jul 20 '17 at 20:12
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There are five Platonic Solids because their definition restricts them to polyhedra.

A Platonic solid is a regular, convex polyhedron. It is constructed by congruent regular polygonal faces with the same number of faces meeting at each vertex.

If all geometric structures are included the result is the twelve FFELLONIC FORMS, a complete series ranging from a triangle to the densest possible space filling honeycomb.

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