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 Course: Seminar Topic: Seminar Description: Proof: Take any Platonic Solid.
• Face is some regular polygon or n-gon with r edges meeting at each vertex. So n>=3 and r>=3.
• Count the Edges: N*F=2E( Because each edges makes 2 faces so each edge is counted twice in n*F).
• Lets take an example of a Cube.
N*F=2E
4*6=2*12 (Proved)
r*V=2E
• By Euler’s Formula V- E + f=2
(2 E / r) – E + (2 E/n ) =2
(n*F = 2E => F =2E/n)
(r*V =2E => V =2E/r)
• Placing the values of F and V)in equation .
After Simplifying we get ((1 / r) +(1/n) =(1/2)+(1/E)).
Now n>=3 and r>=3 but both n and r cannot be >3.(this you can find out by placing the values of n and r greater than 3 in the equation).
• Case 1:
N=3 => (1 / r ) = ( 1/6) + (1/E)
when r =3 E=6 its tetrahedron , r=4 E =12 its octahedron and when r = 5,E=30 its icosahedron.
• Case 2:
R=3 =>(1/n) = (1/6) + ( 1 / E). When n=3,E=6 its tetrahedron=4, E=12 its cube and When n =5 ,E=30 its dodecahedron).
There cannot be any other possible values thus it is proved that there are exactly 5 platonic solids.