# Appendix: Proof of the Gap Theorem

### Gap Theorem

Given a vertex star with exactly two edges that do not have two adjacent polygons, if the sum of the angles of the polygons in the vertex star is s, then define the gap angle (the angle of the polygons needed to complete the vertex star) as

2π - s

and the gap ratio g as:

g = (2π - s)/π

If g = 1 (and so the gap angle is π) then the vertex star can be completed using regular polygons in only three ways: by 3 triangles, 2 squares or a triangle and a hexagon.

If g < 1 (and so the gap angle is < π), then the vertex star can be completed using regular polygons if and only if m = 2/(1-g) is an integer.

In this case, the vertex star can be completed by a single m-sided polygon. There is one additional way to complete the vertex star in only three cases:

If g = 2/3, then the vertex star can be completed by two triangles (in addition to a hexagon).
If g = 5/6, then the vertex star can be completed by a triangle and a square (in addition to a dodecagon).
If g = 14/15, then the vertex star can be completed by a triangle and a pentagon (in addition to a 30-gon).

### Proof

Start by considering the case where g < 1 (and thus the gap angle is < π).

Then at most 2 polygons can fill the gap, because even three triangles, the polygons with the smallest angle, have an angle sum of π and the gap angle must be less than π.

If there are two polygons, then at least one must be a triangle because even two squares, the smallest polygons besides the triangle, have an angle sum of π

In addition to a triangle, the other polygon must have an angle smaller than a hexagon because a triangle and a hexagon have an angle sum of π

Thus, if there are two polygons, the only options are a triangle and a triangle (g = (π/3+π/3)/π = 2/3), a triangle and a square (g = (π/3+2π/4)/π = 5/6) and a triangle and a pentagon (g = (π/3+3π/5)/π = 14/15).

In all three cases, m = 2/(1-g) is an integer.

If the gap can be filled by one m-sided polygon, then

gπ = (m-2)π/m

and solving for m gives

m = 2/(1-g).

Now consider the case where g = 1 and therefore the gap angle is exactly π

If the solution has three or more polygons then it must be three triangles because the sum of the angles of any other three polygons is greater than π.

If the solution has two polygons, then it must consist of polygons with less than 7 sides, because the sum of the angles of a triangle and a heptagon is greater than π

There is no solution with one polygon because the angle of any regular polygon is less than π

A more detailed examination of the options for two polygons shows that two squares and a triangle and a hexagon are the only choices with an angle sum of π.