March 11 2019, 01:42

In the photo is the NASA OV1-8 PASCOMSAT satellite, 1966. This one specifically is inflatable, but engineers, let’s fantasize – what if you were tasked with creating a hollow sphere covered with regular polyhedra, how would you approach the task? Assume it needs to be assembled from pentagons and hexagons. Then, to manufacture pentagons and hexagons separately, one must calculate how many are needed including duplicates, compute the convex and concave features, devise a method for reliable connections, and organize molds for casting. And this is just to manufacture the sphere.

I know the answers to the questions above, just an interesting educational problem for interviews and to assess people’s thinking abilities) And the photo at NASA turned out a bit crooked).

Okay, simple question – how many pentagons are needed to cover a sphere with pentagons and hexagons? The calculation fits into one Facebook comment)

UPDATE: In short, the correct answer is 12. Grigory Bakunov did well) There is a beautiful theorem by Euler, stating that for any convex polyhedron, the number of vertices minus the number of edges plus the number of faces will always equal two. If our polyhedron consists of X hexagons and Y pentagons, then the number of faces will be X+Y, the number of vertices will be (6X+5Y)/3, and the number of edges – (6X+5Y)/2. Here, 5 and 6 represent the number of edges of a pentagon and a hexagon, respectively, and the denominator accounts for the fact that two edges of a flat figure form one edge of a volumetric one, and the same thing for vertices, only with a three. So now, plugging this into Euler’s formula, we find that X drops out of it completely, and Y=12. This means that to tile a sphere with pentagons and hexagons, exactly 12 pentagons are needed, while the number of hexagons can be as many as necessary. Furthermore, to apply this pattern onto a sphere, the flat faces need to be projected onto the spherical surface, where they will simply transform into an arc. To assemble the sphere, you will need 12 pentagons (all identical) and some number of hexagons. Their number depends on the size of the pentagons. This too can be calculated, but it would be rather lengthy.