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<b>Proof: </b>Take any Platonic Solid. <br><ul><li>Face is some regular polygon or ngon with r edges meeting at each vertex. So n>=3 and r>=3. <li>Count the Edges: N*F=2E( Because each edges makes 2 faces so each edge is counted twice in n*F). <li>Lets take an example of a Cube. <br> N*F=2E <br> 4*6=2*12 (Proved) <br> r*V=2E <li>By Euler’s Formula V E + f=2 <br> (2 E / r) – E + (2 E/n ) =2 <br> (n*F = 2E => F =2E/n) <br> (r*V =2E => V =2E/r) <br> <li>Placing the values of F and V)in equation . <br> After Simplifying we get ((1 / r) +(1/n) =(1/2)+(1/E)). <br> 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). <li><b>Case 1:</b> <br> N=3 => (1 / r ) = ( 1/6) + (1/E) <br> when r =3 E=6 its tetrahedron , r=4 E =12 its octahedron and when r = 5,E=30 its icosahedron. <li><b>Case 2:</b> <br> 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). <br>There cannot be any other possible values thus it is proved that there are exactly 5 platonic solids. </ul> <center><img SRC="seminar_41.gif"></center>
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Prof. Ashay Dharwadker