Triangular Tilings of a Sphere

 

Suppose we place an equilateral triangle with edge length 1 on the surface of a sphere of radius R much greater than 1. Now randomly select one of the three edges of this triangle and construct an identical triangle on that edge (also on the surface of the sphere). Now we have four exposed edges, each of which is eligible for the construction of another equilateral triangle. We choose one of these edges at random and construct another triangle. This leaves us with 5 exposed edges, all eligible, from which we randomly select one and construct another equilateral triangle, and so on.

 

Before long there will be some "exposed edges" on which it's not possible to construct an equilateral triangle (without overlapping an existing triangle), so we say those edges are not eligible, and we exclude them from our random selection list. At the beginning of this process our list of eligible edges will obviously increase by 1 at each step, but then it will occasionally not increase on some steps, and eventually it will start to decrease as we continue to adjoin more triangles. Ultimately we will have no more eligible edges.

 

What fraction of the sphere's surface area would we expect to cover in this way? What is the asymptotic behavior of this fraction as R increases? (Since this process would completely tile the plane, it might seem that as R approaches infinity the coverage ratio must approach 1, but perhaps there's a tendency for each local region to "waste" a certain amount of area due to certain patterns of interference that are repeated throughout the surface for any finite R, even though all the interference vanishes when the surface is perfectly flat.) How many ineligible exposed edges would be present at the end? If, instead of randomly deciding where to add the next triangle, we are allowed to plan our moves, is there a simple algorithmic strategy that maximizes the covered area and/or minimizes the final number of ineligible exposed edges?

 

One might think that, since it's possible to fit 5 but not 6 equilateral triangles to a common vertex, the coverage must be around 5/6, regardless of whether the pattern was random or not. However, the coverage may actually be less than that, and it can depend on the sequence in which the tiles are placed. We could ask the opposite question, namely, what is the least number of tiles that leave no eligible edges remaining. Suppose we have a configuration such as the one shown below on the surface of a very large sphere:

 

 

The lines AB and CD are nearly geodesics and essentially parallel as they cross the segment AC. However, they don't remain parallel (as they would on a perfectly flat plane). Instead they are slowly converging, based on the fact that the centerlines of those two rows of tiles must be geodesics, and by symmetry the cross-wise row of tiles along AC must be the point of maximum separation between those geodesics. As a result, not only is it impossible to fill in the 6th edge of the upper "hexagon", it's also impossible to place a triangle anywhere between the lines AB and CD anywhere below C. If we continue this pattern, the lines AB and CD will eventually intersect, but it will take a very long time if the curvature is very slight.

 

Also, we could repeat this pattern, creating a series of "stripes" alternating between tiled and un-tileable regions, as shown below:

 

 

Here the lines EF and GH are also slightly converging, making the region between them unfillable. In this way we might, by design, be able to exhaust our eligible edges while having tiled only about 2/3 of the surface, since the unfillable slice between every two rows of tiles would have about half the area of a row of tiles.

 

Of course, this pattern requires us to carefully arrange our construction. For example, if we didn't tile all the way out to DF on the central path, we could build a tile path from H to B, which wouldn't quite connect perfectly, but would block the central row of tiles. In general, we could construct some very "maze-like" patterns, and if we just randomly selected our sites for each successive triangle, we would (or at least could) get some fairly jagged arrangements with extensive unfillable "slices".  As a result, we'd expect to end up with less than 5/6 coverage... and for it to be possible to get significantly different coverage by design versus random selection.

 

In fact, it's possible to show that we can actually get arbitrarily low coverage simply by constructing one geodesic "belt" around a great circle of the sphere, and then constructing a triangle on each of the exposed faces of that belt. We'll end up with an arrangement that looks like this

 

 

Notice that it's not possible to construct any more triangles on this configuration, since the only exposed edges are subtending an angle just slightly less than what is required to accommodate another triangle. Thus, for an arbitrarily large sphere we have a saturated configuration of tiles that covers just an arbitrarily thin slice around a great circle. Naturally we wouldn't expect to reach this configuration by chance, but it illustrates how this type of tiling can easily become saturated even while large regions of the sphere's surface are totally uncovered.

 

Along these same lines, we could tile along any of the geodesic projections of the edges of a tetrahedron, cube, or icosahedron (whose vertices join three edges), and then "saturate" those geodesic segments by placing triangles on each exposed face, just as we did with the geodesic belt. In this way, we can get just about any amount of coverage, from the max possible all the way down to virtually zero.

 

To achieve coverages close to 1, it might be possible to construct a spiral belt, with "dog legs" as needed in order to circle the sphere and end up just skimming the previous loop, without leaving much empty space. In any case, it's clear that we can have saturated tilings of the sphere with a wide range of coverages, so this again raises the original question, i.e., what is the expected coverage for a random tiling?

 

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