Ponder This Challenge - October 2007 - Resistance of Platonic solids

Ponder This Challenge:

There are 5 platonic solids, the tetrahedron (4 vertices, 4 triangular faces, 6 edges), the cube (8 vertices, 6 square faces, 12 edges), the octahedron (6 vertices, 8 triangular faces, 12 edges), the dodecahedron (20 vertices, 12 pentagonal faces, 30 edges) and the icosahedron (12 vertices, 20 triangular faces, 30 edges). Consider open models of these solids with wire edges connecting the vertices. Suppose each wire has unit resistance. For each case find the total resistance between a pair of adjacent vertices. Express each answer as a rational number.

Hint: The answers can be found by brute force but there is a way to use symmetry.

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Solution

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  Ponder This Solution:

  The answers are:

  • tetraheron -- 1/2* cube -- 7/12* octahedron -- 5/12* dodecahedron -- 19/30* icosahedron -- 11/30

  They can be derived as follows. For any graph the resistance between two points, S and T, can be computed as follows. Set the potential of S to 1, T to 0. Solve a set of linear equations to find the potential of the remaining vertices. Compute the current flow out of S (which will necessarily equal the current flow into T). Invert the current flow to determine the resistance.

  For our case for each graph let n be the number of vertices and d be the degree of each vertex. Specify adjacent vertices S and T. Setting the potential of S to 1 and T to 0 will determine the potential values on the other vertices. Consider rotating the solid around S (so T coincides with each of the neighbors of S). Average the rotated potential values at each vertex. The average at S is clearly 1 (since S remains fixed). The average potential value at each neighbor of S is clearly the same by symmetry. Let this value be x. We claim the average potential at every remaining vertex is also x. This follows because there is no current flow into or out of any of the remaining vertices in any of the rotated states. So what is x? By symmetry the average potential over all the vertices is .5. So since the potential at S is 1 we have x=(n/2 - 1)/(n-1). So the average current out S (which will be the same for each rotation) is d*(1-x)=d*(n/2)/(n-1). So the total resistance between S and T is the inverse of this or (2*n-2)/(d*n) = (2/d)*(1-1/n). Using the values of n and d for each solid we find the answers above.

  Note this method will work for any sufficiently symmetric graph.

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Solvers

• Ashish Srivastava (10.03.2007 @12:37:09 PM EDT)
• V Balakrishnan (10.03.2007 @02:44:54 PM EDT)
• Chris Hills (10.03.2007 @04:05:57 PM EDT)
• Dan Dima (10.03.2007 @04:21:50 PM EDT)
• Jiri Navratil (10.03.2007 @07:13:21 PM EDT)
• Henry Bottomley (10.03.2007 @08:39:27 PM EDT)
• Celia Anteneodo (10.03.2007 @09:57:51 PM EDT)
• John T. Robinson (10.03.2007 @10:52:27 PM EDT)
• Gary M. Gerken (10.03.2007 @11:40:29 PM EDT)
• Abhisek Ukil (10.04.2007 @05:16:10 AM EDT)
• Christian Blatter (10.04.2007 @09:04:17 AM EDT)
• Eugene Vasilchenko (10.04.2007 @10:44:32 AM EDT)
• Frank Yang (10.04.2007 @11:57:10 AM EDT)
• Karl D'Souza (10.04.2007 @08:13:19 PM EDT)
• Parashar Dave (10.05.2007 @08:53:50 AM EDT)
• Andreas Eisele (10.05.2007 @11:55:41 AM EDT)
• Sergio Daher (10.05.2007 @02:53:32 PM EDT)
• Joe Schauder (10.05.2007 @04:17:54 PM EDT)
• Al Zimmermann (10.05.2007 @04:39:10 PM EDT)
• Zhou Guang (10.06.2007 @10:31:56 AM EDT)
• Almudena Moratinos (10.06.2007 @09:38:21 PM EDT)
• Daniel Hsu (10.07.2007 @05:05:04 AM EDT)
• Dion Harmon (10.07.2007 @09:27:16 PM EDT)
• Renato Fonseca (10.07.2007 @10:10:09 PM EDT)
• Arthur Breitman (10.08.2007 @12:30:37 PM EDT)
• Fred Batty (10.08.2007 @05:52:17 PM EDT)
• Alan Murray (10.08.2007 @06:09:31 PM EDT)
• Jose H. Nieto (10.08.2007 @07:42:39 PM EDT)
• Øyvind Grotmol (10.08.2007 @07:55:48 PM EDT)
• Luke Pebody (10.08.2007 @10:22:36 PM EDT)
• Ed Sheppard (10.09.2007 @12:26:15 AM EDT)
• D Li (10.09.2007 @01:18:02 AM EDT)
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• Lukas Saul (10.11.2007 @01:53:59 PM EDT)
• DJCV (10.12.2007 @09:28:40 PM EDT)
• Maxim G. Smith (10.13.2007 @10:48:21 PM EDT)
• Andrea Andenna (10.15.2007 @04:32:23 PM EDT)
• John Ritson (10.15.2007 @05:29:01 PM EDT)
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• Amos Guler (10.16.2007 @04:07:42 PM EDT)
• Nyles Heise (10.17.2007 @08:12:47 PM EDT)
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• Phil Muhm (10.22.2007 @11:44:55 AM EDT)
• Balazs Ugron (10.25.2007 @04:58:16 PM EDT)
• David McQuillan (10.26.2007 @02:09:21 PM EDT)
• Chris Shannon (10.26.2007 @05:56:23 PM EDT)
• Joaquim Neves Carrapa (10.26.2007 @08:55:42 PM EDT)
• Earl Gose (10.27.2007 @04:12:10 AM EDT)
• Didier Bizzarri (10.27.2007 @09:36:21 PM EDT)
• Robert Persch (10.28.2007 @03:19:42 PM EDT)
• Keith F. Lynch (10.28.2007 @05:59:49 PM EDT)
• Daniel Bitin (10.28.2007 @09:33:41 PM EDT)
• Albert Stadler (10.30.2007 @02:47:44 AM EDT)
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• Michael Praehofer (10.31.2007 @04:52:52 AM EDT)
• Karthik Tadinada (10.31.2007 @02:37:49 PM EDT)
• Michael Slifker (10.31.2007 @02:51:36 PM EDT)
• Rogerio Ponce da Silva (11.01.2007 @04:00:22 AM EDT)