1 problem found
A polyhedron is a solid bounded by \(F\) plane faces, which meet in \(E\) edges and \(V\) vertices. You may assume \textit{Euler's formula}, that \(V-E+F=2\). In a regular polyhedron the faces are equal regular \(m\)-sided polygons, \(n\) of which meet at each vertex. Show that $$ F={4n\over h} \,, $$ where \(h=4-(n-2)(m-2)\). By considering the possible values of \(h\), or otherwise, prove that there are only five regular polyhedra, and find \(V\), \(E\) and \(F\) for each.
Solution: Note that each of the \(F\) faces have \(m\) edges. If we count edges from each face we will get \(mF\) edges, but this counts each edge twice, so \(E = \frac{m}{2}F\). Similarly each edge goes into \(2\) vertices, but this counts each vertex \(n\) times, therefore \(V = \frac{2E}{n} = \frac{m}{n}F\) Therefore \begin{align*} && 2 &= V - E + F \\ &&&= \frac{m}{n}F - \frac{m}{2}F + F \\ &&&= F \left ( \frac{2m-nm+2n}{2n} \right) \\ &&&= F \left ( \frac{4-(n-2)(m-2)}{2n} \right) \\ \Rightarrow && F &= \frac{4n}{4-(n-2)(m-2)} = \frac{4n}{h} \end{align*} Notice that \(1 \leq h \leq 4\), so If \(h = 4\), then \(n = 2\) or \(m = 2\) but we can't have two-sided polygons or polyhedra with two faces making a vertex so this is not possible. If \(h = 3\) then \((n-2)(m-2) = 1\) and \(3 \mid n\), so we have \(n = 3, m = 3\). ie we have triangular faces meeting at \(3\) per point. We also have \(F = \frac{4 \cdot 3 }3\) which is \(4\) faces so this is a tetrahedron and \((V,E, F) = (4, 6, 4)\) If \(h = 2\) then \((n-2)(m-2) = 2\): Case 1: \(n = 4, m = 3\) which is four triangles meeting at each vertex with \((V,E,F) = (6, 12, 8)\), ie an octohedron. Case 2: \(n = 3, m = 4\) so we have three squares meeting at each vertex, with \((V,E,F) = (8,12,6)\) which is a cube. If \(h = 1\) then \((n-2)(m-2) = 3\) Case 1: \(n = 5, m = 3\) which is five triangles meeting at each vertex, with \((V,E,F) = ( 12,30, 20)\) ie an icosahedron. Case 2: \(n = 3, m = 5\) which is three pentagons meeting at each vertex, with \((V,E,F) = (20, 30,12)\) which is a dodecahedron. These are all the possible cases and hence we have found the five platonic solids.