EDGE TESSELLATIONS AND THE STAMP FOLDINGPROBLEMMATTHEW KIRBY AND RON UMBLEAbstract. We prove that a polygon generating a tessellation of the planewhen re‡ected in its edges is either a regular hexagon; a 60-90-120 kite; a120-rhombu s; a rectangle; an equilateral, a 30-right, an isosceles right, or a120-isosceles tri angle. Of these, the four whose interior angles measure atmost 90solve the “Stamp Folding P roblem.”How many ways can one con …gure the perforation lines on a she et of postagestamps so that it folds conveniently into a stack of single stamps? The solution ofthis “Stamp Folding Problem” follows from our main result, which was discoveredby Millersville University students Andrew Hall, Joshua York, and the …rst authorduring an undergraduate research seminar directed by the second author in thespring of 2009: Theorem 1. A polygon that generates a tessellation of the planewhen re‡ected in its edges is either a regular hexagon; a 60-90-120 kite; a 120-rhombus; a rectangle; an equilateral, a 30-right, an isosceles right, or a 120-isoscelestriangle.A tessellat ion (or tiling) of the plane is a collection of plane …gures that …lls theplane with no overlaps and no gaps. An edge tessellation is generated by re‡ectinga polygon in its edges. Edge tessellations are the most symmetric examples of Lavestilings, as one can see from the complete list in [3]. The name “Laves tiling”honorsthe German crystallographer Fritz Laves (1906-1978).Corollary 1. A polygon generating an edge tessellation that solves the Stamp Fold-ing Problem is either a rectangle; an equilateral, a 30-right, or an isosceles righttriangle.Figure 1. Edge tessellations with interior angles at most 90:Date: April 15, 2010.Key words and phrases. Symmetry, tessellation, wallpaper pattern .12 MATTHEW KIRBY AND RON UM BLEOn page 143 of his book Piano-Hinged Dissections: Time to Fold! [2], G. Fred-erickson conjectures that the various types of triangles in Figure 1 are the only onesthat “seem suitable for [stamp] folding puzzles.” To prove this fact, observe thatperforation lines in a sheet of postage stamps that folds into a stack of single stampsare lines of symmetry missing the stamps’ interiors. Consequently, the polygonsin Theorem 1 whose interior angles measure at most 90generate solutions of theStamp Folding Problem (se e Figure 1). The other four polygons in Theorem 1fail to generate solutions because each has an interior angle of 120whose bisectorcontains an edge of some stamp (see Figure 6).Theorem 1 is motivated by recent work of A. Baxter and the second author, whouse an edge tessellation to …nd, classify, and count equivalence classes of periodic or-bits on equilateral triangles [1]. Indeed, other polygons to which their classi…cationtechnique can be applied necessarily generate an edge tessellation. Thus Theorem1 suggests potentially seven projects for the interested reader: Find, classify andcount classes of periodic orbits on the seven other polygons in Theorem 1.A partial resu lt was obtained by B. Maskit in [5], who showed that polygonswith interior angle measures in the set S0= fx  90j nx = 180; n 2 Ng generatean edge tessellation. This condition is su¢ cient but not necessary, as it misses thepolygons in Theorem 1 with an interior angle of 120: One is tempted to simplyadjoin the element 120to S0, but doing so admits a 60-90-90-120 trapezoid, forexample, so the solution is more subtle. Let us proceed with the proof of Theorem1.Let G be a polygon that generates an edge tessellation, let V be a vertex of G;and let  < 180be the measure of the interior angle at V: Let G0be the image ofG when re‡ected in an edge of G containing V: Then the interior angle of G0at Vhas measure , and inductively, the interior angle at V in every copy of G sharingvertex V has measure  (see Figure 2).Figure 2. Congruent interior angles at vertex V:Since successively re‡ecting in the edges of G that meet at V is a rotationalsymmetry, V is the center of a group of n rotational symmetries with minimumpositive rotation angle n= 360=n and  2 f180=n; 360=ng ; thus we refer to Vas an n-center. If n = 2; then 2= 180and m\V = 90; note that the bisector of\V is not a line of symmetry.The case n = 3 is especially important. Since 3= 120and n 90whenevern  4; an obtuse angle of G measures 120: Thus if V is a vertex of G; the measureof the interior angle at V lies in the setS = fx  120j nx = 360; n 2 Ng=n120; 90; 72; 60; 5137; 45; 40; 36; : : : ; 18; : : :o:EDGE TESSELLATIONS AND THE STAM P FOLDING PROBLEM 3Now suppose that m\V = 120; then three copies of G share the vertex V: Let eand e0be the edges of G that meet at V; and labeled so that the angle from edgee to edge e0measures 120(see Figure 3). Let e0and e00be their resp ective imagesunder a 120rotation. Then e00lies on the bisector of \V and is the re‡ection ofe0in e: By a similar argument, if an odd number of copies of G share vertex V; thebisector of \V is a line of symmetry.Figure 3. A line of symmetry b isects a 120interior angle.Let g be the number of edges of G: Then the interior angle sum 180(g  2) 120g implies g  6: If g = 5; the interior angle sum of 540implies that G has atleast one vertex V such that m\V = 120: Thus G is symmetric with respect tothe angle bisector at V and the other interior angles of G pair o¤ congruently–twowith measure x; two with measure y: Note that the interior angles in one of thesepairs are adjacent (see Figure 4). If x = y; then x = 105=2 S; hence x 6= y: Ifx < y; then y > 90and the only possibility in S is y = 120: But if three interiorangles of G are bisected by lines of symmetry, then x = y; which is a contradiction.Therefore g 6= 5 and G is a triangle, a quadrilateral, or a hexagon.Figure 4. A pentagon G with an interior angle of 120at V .If G is a h exagon, each interior angle i 120for 1  i  6: Thus for eachi; there exists i 0such that i+ i= 120: But 6 (120) =P(i+ i) =Pi+Pi= 720+PiimpliesPi= 0so that i= 0for all i: Hence Gis equiangular and symmetric with respect to each of its interior angle bisectors;therefore G is a regular hexagon.Now suppose that G is a quadrilateral or a triangle with an interior angle of120. If G is a triangle, it is 120-isosceles by symmetry, so