Question: A polyhedron (3-polytope) is called regular if all facets are congruent regular polygons and all the angles at the vertices are equal. Supply the details in following proof that there are only five regular polyhedral.

a. Suppose that regular polyhedra has r facets, each of which is a k side regular polygon, and then s edges meet at each vertex. Letting v and e denote the numbers of vertices and edges in the polyhedron, explain why \(kr = {\bf{2}}e\) and \(sv = {\bf{2}}e\).

b. Use Euler’s formula to show that \(\frac{{\bf{1}}}{s} + \frac{{\bf{1}}}{k} = \frac{{\bf{1}}}{{\bf{2}}} + \frac{{\bf{1}}}{e}\).

c. Find all the integral solutions of the equation in part (b) that satisfy the geometric constraints of the problem. (How small can k and s be?)

For your information, the five regular polyhedra are the
tetrahedron (4, 6, 4), the cube (8, 12, 6), the octahedron (6, 12,
8), the dodecahedron (20, 30, 12), and the icosahedron (12,
30, 20). (The numbers in parentheses indicate the numbers of
vertices, edges, and faces, respectively.)

As each edge belongs to two facets, therefore \(kr\) is twice the number of edges, so \(kr = 2e\).

Also, there are two vertices on each edge, therefore \(sv = 2e\).

According toEuler’s formula \(v - e + r = 2\).

Substitute \(v = \frac{{2e}}{s}\) and \(r = \frac{{2e}}{k}\) in the Euler’s formula.

\(\begin{array}{c}\frac{{2e}}{s} - e + \frac{{2e}}{k} = 2\\2e\left( {\frac{1}{s} - \frac{1}{2} + \frac{1}{k}} \right) = 2\\\frac{1}{s} + \frac{1}{k} = \frac{1}{e} + \frac{1}{k}\end{array}\)

Hence proved.

A polygon has atleast 3 sides \(\left( {k \ge 3} \right)\) and 3 vertices \(\left( {s \ge 3} \right)\).

But from the equation \(\frac{1}{s} + \frac{1}{k} = \frac{1}{e} + \frac{1}{k}\), both s and k cannot be equal to 3.

So, for \(k = 3\) it is:

\(\frac{1}{s} - \frac{1}{6} = \frac{1}{e}\)

From the above equation, for \(s = 3,\,4\,{\rm{or}}\,5\), \(e = 6,\,12\,\,{\rm{or}}\,30\), corresponding to tetrahedron, octahedron, and icosahedron.

For \(s = 3\) it is:

\(\frac{1}{k} - \frac{1}{6} = \frac{1}{e}\) for \(k = 3,\,4\,{\rm{or}}\,5\), \(e = 6,\,12\,\,{\rm{or}}\,30\), corresponding to tetrahedron, cube, and dodecahedron.