We call a convex polyhedron a sandpile if one of the faces, called base, has a common edge with all the other faces. Furthermore, each vertex is linked with three edges. We consider the sandpiles by vertex translations, rotations and symmetries close. Therefore, there are 3 sandpiles with a hexagonal base:

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How many sandpiles have a base with 9 edges?

  • $\begingroup$ By some graph theory, I know there are going to be 7 internal vertices in each of your sandpiles with a 9-gon base. $\endgroup$ – Brian Hopkins Jul 25 '20 at 1:57


This is dual to triangulations of a 9-gon that, without symmetries, are counted by a Catalan number. Taking symmetry into account leaves the sequence A000207 first considered by Motzkin: starting from a triangle base, it's 1, 1, 1, 3, 4, 12, 27, 82...

For the duality, consider the hexagon examples. Build a triangulated hexagon from each one in the following way: The vertices of the new hexagon are are midpoints of the base hexagon edges. Connect these new vertices if their corresponding faces share an edge. The top left figure becomes a hexagon containing an N, the top right has three internal edges emanating from the left vertex, and the bottom figure has a triangle inside. In this MathWorld entry, in the two rows of 7 triangulated hexagons, these are row 2 #6, row 1 #7, and row 2 #4, respectively. As you can see from that list, the other 11 triangulations are equivalent to one of these three.


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