The n-Category Café

The photo and the diagram do not match.

Here’s a way to argue that Gerard’s solution to my puzzle is correct. Thurston showed that any star of the sort I described folds up so that all 11 of its tips touch and you get a convex polyhedron. Thus, this diagram must give some convex polyhedron:

and we just need to show it’s the regular icosahedron.

To do this, it’s enough to show it’s a polyhedron with equilateral triangular faces, with 12 vertices, and 5 triangles meeting at each vertex.

Here’s a rough argument, which would need to be improved to become a proof.

1) First let’s show that — like all Thurston polyhedra — this polyhedron has 12 points with 5 yellow triangles meeting at them.

That’s easy to see for the 11 vertices that arise from the 11 ‘indentations’ of the star. Each of those indentations arises from cutting out a green equilateral triangle, so when we fold up our star each indentation gives a vertex with an angle deficit of π/3. Thus, 5 small equilateral triangles must meet there: one short of the 6 you’d need to eliminate the angle deficit.

But what about the 12th vertex of our polyhedron, the one where all the tips of our star meet when we fold it up? Why do 5 small equilateral triangles meet there?

We could just look and think about it. But it’s easier to invoke Gauss–Bonnet (or Euler’s formula) which implies that the sum of the angle deficits at the vertices of a convex polyhedron is always $4 \pi$. Since we’ve already seen 11 vertices of our polyhedron have an angle deficit of $\pi/3$, the 12th must have an angle deficit of

$$4\pi - 11 \cdot \frac{\pi}{3} = \frac{12 - 11}{3} \pi = \frac{\pi}{3}$$

as well. So 5 small equilateral triangles must meet there too!

Then let’s note that in this polyhedron–unlike all the other Thurston polyhedra—there are no points at which 6 triangles meet. There are none visible from the start in our star, and none get formed when we fold it up. For comparison, in the example of the hexagonal antiprism, there is one such point visible from the start, and 1 more gets formed when we fold it up. Gerard has marked them both:

So, I argue, the polyhedron we get has only equilateral triangular faces, and 12 vertices at which 5 faces meet. Thus, by known facts, it’s the regular icosahedron.

(It’s fun to look for soft spots in this argument.)

Here is another interesting example found by Gerard Westendorp. It’s part of an infinite family: