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 :-)

Now go visit my blog please, or look at other interesting maths stuff :-)

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