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# Shape-density of a Dodecahedron

Suppose we have a regular polyhedron with side length 1 rolling on an infinite sheet of paper. Every time it rolls on a new face, the outline of the face touching the paper is drawn in pen. This is done continuously until no more new pen marks can be made on the paper. Now, define the "shape-density" of a particular polyhedron to be the average area of every single shape drawn in the plane after the process described above, where no two shapes overlap.

Obviously, the "shape-density" of a cube is $$1$$, because after the process we get a grid lattice. In addition, the "shape-densities" of a tetrahedron, octahedron and icosahedron are all $$\dfrac{\sqrt{3}}{4}$$, because we get a triangular lattice after the process. What is the "shape-density" of a dodecahedron?

Please tell me if I am not making any sense.

Note by Daniel Liu
2 years, 3 months ago

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Liu, I'm going to guess that what you are talking about involves plotting all the points on the plane that can be "reached" via rolling around a dodecahedron with unit edges, the vertices of it landing on such points. That would be the same as attempting to plot all points with coordinates

$$\left(\displaystyle \sum _{ n=1 }^{ 10 }{ { A }_{ n }Cos\left( 36n \right) } ,\displaystyle \sum _{ n=1 }^{ 10 }{ { A }_{ n }Sin\left( 36n \right) } \right)$$

where $${A}_{n}$$ are $$10$$ integers of any size. If we just consider the values of $$Cos\left(36n \right)$$, then it's not hard to see that we could be looking at values $$a+b\sqrt { 5 }$$ where $$a$$ and $$b$$ are positive or negative integers of arbitrary size, which means we can make it as small as we want. More work with similar thinking shows that we can "reach" a point as close to the origin, or any other point, as we'd like (the proof is not trivial). That means, effectively, such points can be everywhere dense on the plane, so that the "shape-density" would approach $$0$$. At least that's how I understand the wording of your problem. Maybe you can help clarify?

Note that in the case of the triangular lattice, we don't have the form $$a+b\sqrt{3}$$, so that we don't run into the same problem. · 2 years, 3 months ago

No, your interpretation is correct. I myself didn't know the answer to the problem, which is why I posted it. Thanks for your solution outline; I did not expect the answer to approach $$0$$. · 2 years, 3 months ago