Why only 5 Plato solids ?
Plato Solid is: Regular Polyhedron 正多面体
- Each Face is n-sided polygon
- Each Vertex is common to m-edges (m ≥ 3)
Only 5 solids possible:
Tetrahedron (n,m)=(3,3) 正四面体
Hexahedron (or Cube) (n,m)=(4,3) 正六面体
Octahedron (n,m)=(3,4)正八面体
Dodecahedron (n,m)=(5,3)正十二面体
Icosahedron (n,m)=(3,5)正二十面体
Proof:
Since each Edge (E) is common to 2 Faces (F)
=> n Faces counts double the edges
nF = 2E …(1)
Since each Vertex has m Edges, each Edge has 2 end-points (Vertex).
=> m Vertex counts double the edges
mV = 2E …(2)
(1) : E= n/2 F
(2): V= 2/m. E = n/m. F
(1) & (2) into Euler Formula: V -E + F = 2
(n/m. F) – (n/2.F) + F = 2
F.(2m + 2n – mn) = 4m
Since F>0 , m>0
=> (2m + 2n – mn) >0
=> – (mn -2n -2m) >…
=> (mn -2n -2m) <…
=>…
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