## Icosahedron’s Resistance

I rarely post physics puzzles, but this one is too good to pass on.

Puzzle. A wireframe icosahedron is assembled so that each of its edges has a resistance of 1. What is the total resistance between opposite vertices of the icosahedron?

While we are at it, another interesting question would be the following.

Puzzle. A wireframe cube is assembled so that each of its edges has a resistance of 1. What is the total resistance between opposite vertices of the cube?

And this reminds me of a question I heard when I was preparing for an IMO many years ago.

Puzzle. A wireframe infinite square grid is assembled so that each of its edges has a resistance of 1. What is the total resistance between two neighboring vertices?

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1. #### Lazar Ilic:

Platonic Relationships Among Resistors by Bradley Allen and Tongtian Liu is a solid introduction.

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5/6

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2. #### Sanandan Swaminathan:

Icosahedron: It has 20 triangular faces with 5 edges meeting at each vertex. Let A and B be a pair of super-diagonally opposite vertices. Let A be connected to vertices a1, a2, a3, a4, and a5. Let B be connected to vertices b1, b2, b3, b4, and b5. The two sets {a1, a2, a3, a4, a5} and {b1, b2, b3, b4, b5} have no vertex in common. Connect a battery across vertices A and B. Assume current i exits from A, through the battery, and enters at B. By symmetry, current i/5 will flow from each of the vertices a1, a2, a3, a4, a5 to A. Similarly, current i/5 will flow from B to each of the vertices b1,b2,b3,b4,b5. By symmetry, a1, a2, a3, a4, a5 will be at the same potential, say V1. So, no current will flow between those vertices. Similarly, b1, b2, b3, b4, b5 will be at the same potential, say V2. So, no current will flow between those vertices.

Let b1 be connected to vertices b2,b5,a1,a5 (apart from B). The current i/5 that flows from B to b1 will get split as i/10 from b1 to a1 and i/10 from b1 to a5 (with no current flowing from b1 to b2 or b5 since b1,b2,b5 are equi-potential points). Consider the path B -> b1 -> a1 -> A. Given that each resistor is 1 ohm, the potential drop across this path is (i/5)*1 + (i/10)*1 + (i/5)*1 = i/2. If the total resistance between A and B is R (the wireframe replaced by a single resistor of R ohms between A and B), then the potential drop between B and A would be i*R since current i flows through this simplified circuit. Thus, i*R = i/2, so R = 1/2 ohm.

Cube: Let the vertices of the top face be A,B,C,D in clockwise direction. Let the vertices of the bottom face be E,F,G,H in clockwise direction, with E directly below A. Let A and G be the super-diagonally opposite vertices across which we need to find the total resistance. Connect a battery across vertices A and G. Assume current i flows from the battery into vertex A and flows out of vertex G into the battery. By symmetry, current i/3 will flow from A to each of its neighboring vertices B,D,E. Similarly, i/3 will flow into G from each of C,F,H.

Current i/3 flowed into vertex B. Vertex G is symmetrically positioned from B’s perspective, so current i/6 will flow from B to C, and i/6 from B to F. Consider the path A -> B -> C -> G. Given that each resistor is 1 ohm, the potential drop across this path is (i/3)*1 + (i/6)*1 + (i/3)*1 = 5i/6. If the total resistance between A and G is R (the wireframe replaced by a single resistor of R ohms between A and G), then the potential drop between A and G would be i*R since current I flows through this simplified circuit. Thus, i*R = 5i/6, so R = 5/6 ohm.

Infinite square grid of resistors: Consider some two adjacent vertices A and B. Connect a current source S1 that sends current i into vertex A, and another identical current source S2 connected to B with opposite polarity (current i exits from vertex B). The other ends of both identical current sources are grounded. We can use the superposition theorem by disconnecting one current source at a time. First disconnect S1. By symmetry, we would have i/4 current flowing towards vertex B through each of the 4 resistors connected to B. We would have i/4 flowing from A to B. Next, disconnect S2 and connect S1. By symmetry, we would have i/4 current flowing from vertex A through each of the 4 resistors connected to A. We would have i/4 flowing from A to B. By the superposition theorem, if we were now to connect both current sources, we would have i/4 + i/4 = i/2 flowing from A to B. With a 1 ohm resistor between A and B, the potential drop from A to B would be (i/2)*1 = i/2. If the total resistance between A and B is R (the wireframe replaced by a single resistor of R ohms between A and B), then the potential drop between A and B would be i*R since current i flows through this simplified circuit. Thus, i*R = i/2, so R = 1/2 ohm. We can save money – instead of buying an infinite number of 1 ohm resistors and taking the resistance across some pair of adjacent vertices, buy just twenty 1 ohm resistors, make an icosahedron wireframe with the 20 resistors, and take the resistance across a pair of super-diagonally opposite vertices ;).

3. #### Sanandan Swaminathan:

Inconsequential typo in the last sentence in my post above. Icosahedron has 30 edges, so needs 30 resistors, not 20.