Polygons That Generate an Edge Tessellation of the Plane

An edge tessellation of the plane is generated by the reflection of some polygon in its edges. We prove that a polygon generating an edge tessellation is exactly one of the following: An equilateral, 30-right, isoceles right, or 120-isoceles triangle, a rectangle, a 60-rhombus, a 60-90-120-kite, or a regular hexagon. The results in this paper were discovered by Millersville University students Andrew Hall, Matthew Kirby, and Joshua York during an undergraduate research seminar directed by the fourth author in the spring of 2009. The goal of the seminar was to identify those polygons that generate a tessellation of the plane when reflected in their edges. Such tessellations, referred to here as edge tessellations, are examples of Laves tilings [3]. This problem is motivated by recent work by A. Baxter and R. Umble who use an edge tessellation to find, classify, and count classes of periodic orbits on equilateral triangles [1]. Indeed, a polygon to which their classification techniques can be applied necessarily generates an edge tessellation. Consider a polygon G that generates an edge tessellation. Observe that the measures of the interior angles of G lie in the set {30◦, 45◦, 60◦, 90◦, 120◦} . Let θ be the measure of the interior angle at some vertex V of G, and note that θ < 180. If G is the reflection of G in an edge containing V , the interior angle of G at V has measure θ, and inductively, every interior angle at V has measure θ. Thus V is a point of n-fold symmetry since successively reflecting in the edges of G that meet at V is a rotational symmetry. But n = 2, 3, 4, 6 by Crystallographic Restriction [4]. Consequently, if G is the image of G under a rotational symmetry, the minimum rotation angle is θ and V is shared by 3, 4, or 6 copies of G. Note that when n = 3, exactly three copies of G share vertex V and G is symmetric with respect to its angle bisector at V . Otherwise, G and G rotate together as a unit, in which case the minimum rotation angle is 2θ and V is shared by 4, 6, 8, or 12 copies of G. The conclusion now follows. Next, let e be the number of edges of G; then the interior angle sum S = 180 (e− 2) . On the other hand, S ≤ 120e since the interior angles measure at most 120. But 180 (e− 2) ≤ 120e implies e ≤ 6 and G has at most 6 edges. Furthermore, it is a well-known fact that there are no regular tilings of the plane by pentagons [4]. Therefore G is a triangle, a quadrilateral, or a hexagon. The fact that G has 3, 4, or 6 edges, has interior angles measuring 30, 45, 60, 90◦, or 120◦, and is symmetric with respect to its angle bisectors at interior angles of 120 proves: Date: July 10, 2009.