Proof of Pick’s Theorem for Three Dimensions

If a polyhedron is formed on a unit grid such that all of its sides are on one of the unit grid segments, all of its border sides connect to exactly two faces, there are no topological holes or hollow parts, and there are no curved faces, then the volume of that polyhedron can be determined by

V = ⅛(b_e + 4i_e – 4i_p – 4)

where b_e is the number of edges on the border of the polyhedron, i_e is the number of edges on the interior of the polyhedron, and i_p is the number of points on the interior of the polyhedron.

Since the first restriction is that all sides are on one of the unit grid segments, this means that all eligible polyhedrons must be made out of unit blocks.  So to prove this variation of Pick’s Theorem for three dimensions, the first step is to show that it is true for a single block, and the second step is to show that combining a unit block to another eligible polyhedron made up of unit blocks preserves this theorem.

                                            

The first step is to show that this variation of Pick’s Theorem for three dimensions is true for a single block.  This can be verified easily enough.  A single block has 12 border edges, 0 interior edges, 0 interior points, and 0 holes.  Therefore, using the formula we can calculate its volume to be V = ⅛(b_e + 4i_e – 4i_p + 4h – 4) = ⅛(12 + 4·0 – 4·0 + 4·0 – 4) = 1, which is exactly the volume we would expect.

The second step is to show that combining a unit block to another eligible polyhedron made up of unit blocks preserves this variation of Pick’s Theorem.  Depending on the already existing polyhedron, the unit block can be combined so that it shares 1 face, 2 faces (adjacently), 3 faces (adjacently in a corner or adjacently in a row), 4 faces (adjacently in a corner), or 5 faces.  As the table below shows, all of these combinations keep the equation to the theorem balanced:

1 shared face	8 new border edges: beN = beO + 8 0 new interior edges: ieN = ieO 0 new interior points: ipN = ipO volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) (VO+1) = ⅛((beO+8)+4ieO–4ipO–4) VN = ⅛(beN+4ieN–4ipN–4)
2 shared  adjacent faces	5 – 1 = 4 new border edges: beN = beO + 4 1 new interior edge: ieN = ieO + 1 0 new interior points: ipN = ipO volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(beO+4ieO–4ipO–4+4+4) VO+1 = ⅛(beO+4+4ieO+4–4ipO–4) VO+1 = ⅛((beO+4)+4(ieO+1)–4ipO–4) VN = ⅛(beN+4ieN–4ipN–4)
3 shared adjacent faces in a corner	3 – 3 = 0 new border edges: beN = beO 3 new interior edges: ieN = ieO + 3 1 new interior point: ipN = ipO + 1 volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(beO+4ieO–4ipO–4+12–4) VO+1 = ⅛(beO+4ieO+12–4ipO–4–4) VO+1 = ⅛(beO+4(ieO+3)–4(ipO+1)–4) VN = ⅛(beN+4ieN–4ipN–4)
3 shared  adjacent faces in a row	2 – 2 = 0 new border edges: beN = beO 2 new interior edges: ieN = ieO + 2 0 new interior points: ipN = ipO volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(be+4ieO+8–4ipO–4) VO+1 = ⅛(beO+4+4(ieO+2)–4ipO+4(hO+1)–4) VN = ⅛(beN+4ieN–4ipN–4)
4 shared  adjacent faces in a corner	1 – 5 = 4 less border edges: beN = beO – 4 5 new interior edges: ieN = ieO + 5 2 new interior points: ipN = ipO + 2 volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(beO+4ieO–4ipO–4–4+20–8) VO+1 = ⅛(beO–4+4ieO+20–4ipO–8–4) VO+1 = ⅛(beO–4+4(ieO+5)–4(ipO+2)–4) VN = ⅛(beN+4ieN–4ipN–4)
5 shared faces	8 less border edges: beN = beO – 8 8 new interior edges: ieN = ieO + 8 4 new interior points: ipN = ipO + 4 volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(beO+4ieO–4ipO–4–8+32–16) VO+1 = ⅛(beO–8+4ieO+32–4ipO–16–4) VO+1 = ⅛(beO–8+4(ieO+8)–4(ipO–4)–4) VN = ⅛(beN+4ieN–4ipN–4)

Note: the following combinations are disregarded since holes or hollowed parts are restricted: 2 opposite faces (because a hole would be formed), 4 adjacent faces in a row (because it would start with a hole), and 6 faces (because it would start with a hollow part).

Since we have shown that this variation of Pick’s Theorem for three dimensions is true for a unit block, and have shown that this variation of Pick’s Theorem is preserved for combining a unit block to another eligible polyhedron made up of unit blocks, this variation of Pick’s Theorem must be true for all polyhedrons as well.