Proof that there are only 5 Platonic Solids

© Stu Savory, 2005.

Got a reply from Ranjit, replying to tuesday's blog-entry, asking "How do you know there are only five regular polyhedra (= Platonic Solids)? Can you prove it? There might be another really big regular polyhedron with lots of corners, edges and faces that nobody has discovered yet!"

Well, sorry to disppoint you, Ranjit, but there are only five; I've appended a simple proof below. Non-mathematically minded blogreaders may stop reading here ;-)

For some X, each face of a regular polyhedron is a regular Y-agon.
At each of the C corners of the regular polyhedron, X edges meet, for some X.
C corners times X edges per corner imply that there are E = C*X/2 edges to the regular polyhedron (the half taking into account that each edge at a corner is counted twice, once at each end).

Each face F has Y edges therefore F = 2*E/Y (the factor 2 taking into account that each edge is where 2 faces meet).

Now we can use Euler's formula we already saw in tuesday's blog-entry : C - E + F = 2 , and substitute 2*E/Y for F and write C*X/2 for E. Simplifying and rearranging, we get :-

2 = (C/2Y)*(2Y-X(Y-2)) = (C/2Y)*{4-(Y-2)(X-2)}

Now it is obvious that both 2 and C/2Y are positive, thus 4-(Y-2)(X-2) must be positive too.

But this implies that 4 > (Y-2)*(X-2), and since X and Y are positive integers with X>=3 and Y>=3, only the following combinations for X and Y are possible :- (Y,X)=(3,3),(3,4),(3,5),(4,3) and (5,3). Five solutions = 5 regular polyhedra, Q.E.D!

Now we can plug these 5 pairs back into the substitutions to get C,F,E values:-

(3,3) 4,4,6 Tetrahedron
(3,4) 6,8,12 Octahedron
(3,5) 12,20,30 Icosahedron
(4,3) 8,6,12 Cube
(5,3) 20,12,30 Dodecahedron

All clear now, Ranjit? You see there are only five, none as yet undiscovered :-)

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