Tessellating With Regular Polygons

Tessellating with Regular Convex Polygons (Supplement) by Justine Leon A. Uro e-mail: [email protected] Section 1. Introduction It is well-known that the equilateral triangle, the square, and the regular hexagon are regular polygons each of which, by itself, can tile the plane completely. It is also known (although not as well-known as the latter) that these ones are the only regular polygons that can do so. Examples of these tessellations are given in Figures 1 to 5. Figures 4 and 5 are called “slide tessellations” since they can be obtained from Figures 1 and 2, respectively, by sliding the appropriate polygons along the edges of the other polygons. Figure 1

Figure 2

Figure 3

Figure 4

Figure 5

Observe that for a regular polygon to completely tile the plane, at least one of the following must be true of the vertices of the polygons in the tessellation. Case I: V vertices of such regular polygons meet about a point (a vertex) without gaps or overlaps as shown in Figures 1 to 3; or, Case II: U vertices of such regular polygons meet at a point (not a vertex) lying on an edge of one such polygon so as to fill, without gaps and overlaps, the space to one side of an edge of one such polygon (e.g., the vertices in the slide tessellations shown in Figures 4 and 5). Hence, Cases I and II, are good sources of regular polygons that may tile the plane completely. Case I is equivalent to saying that the interior angle of the regular polygon divides 360° while Case II is equivalent to saying that the interior angle of the regular polygon must divide 180°. To therefore prove the proposition that among the regular n-gons only the equilateral triangle, the square, and the regular hexagon can tile the plane completely, one may proceed in the following manner: a) determine all those regular polygons with interior angles less than 360° ; then, b) determine which among those in a) have interior angles that divide 360° or 180° ; finally, c) demonstrate which among those in b) can tile the plane completely. In a previous article (see [1]), by this same author, he presented a partial proof of this proposition, the approach of the proof being different from the one just described. This proof was based on a method for finding the solutions to the Diophantine equation x + y = xy which (the method) can be found in [2]. The proof, however, only considers the possibilities described under Case I above (i.e., tessellations shown in Figures 1 to 3) and does not take into account the slide tessellations shown in Figures 4 and 5. Furthermore, it is not explained clearly why each of the three regular polygons previously mentioned can tile the plane completely. In this article, we present the missing details of the proof presented by the author in [1]. More precisely, in Section 2 of this article, the author presents a proof for the possibilities shown in Figures 4 and 5 (i.e., the possibilities described under Case II). In Section 3, the author demonstrates that these three regular polygons can each tile the plane completely. This is done using two methods which the author calls “translated induction” and “magnified (or geometric) induction.”

Section 2. Proof of Case II Consider the following proposition. Proposition 1: If a regular polygon tessellates the plane completely then it is either an equilateral triangle, a square, or a regular hexagon. Proof: Suppose a regular n-gon tessellates the plane completely. Then either Case I or Case II in Section I must be true. Note that each of the interior angle of the n-gon measures (180° - 360°/n). For the proof of Case I, see [1]. For the proof of Case II, suppose U vertices meet at a point lying on an edge of one such polygon so as to fill, without gaps and overlaps, the space to one side of an edge of the polygon (e.g., the vertices in the tessellations in Figures 4 and 5). Then, (number of coincident vertices to one side of an edge) x (measure of an interior angle) = 180° U(180° - 360°/n) = 180° U(1 - 2/n) = 1 U(n - 2) = n U(n - 2) - n = 0 U(n - 2) - (n - 2) = 2 (U – 1)(n – 2) = 2. Solving this in the manner presented in [1] yields Table 1. Possible solutions to (U – 1)(n – 2) = 2. no. of vertices no. of edges Case no. U-1= n-2= U= n= 1

2

1

3

3

2

1

2

2

4

3

-2

-1

-1

1

4

-1

-2

0

0

Cases 3 and 4 are absurd. Case 1 corresponds to the tessellation in Figure 4. Three vertices (U = 3) of three equilateral triangles (n = 3) fill, without gaps and overlaps, the space to one side of an edge of another equilateral triangle in the tessellation. Case 2 corresponds to the tessellation in Figure 5. Two vertices (U = 2) of two squares (n = 4) fill, without gaps and overlaps, the space to one side of an edge of another square in the tessellation. Hence, the equilateral triangle and the square may tessellate the plane completely in this manner, and if they can do so, are the only regular polygons that can produce slide tessellations. This proof of Case II together with the proof of Case I completes the proof of Proposition 1.

Section 3. Constructing the Tessellation One can easily demonstrate that the equilateral triangle, the square, and the regular hexagon can each tessellate the plane completely. We first demonstrate this for the square. A possible method which works only for the equilateral triangle and the square, is what the author calls magnified or geometric induction. We first demonstrate this for the square. Start with the basic unit square or “center square” for the tessellation. The induction step proceeds in the following manner. About this initial unit square, construct eight squares congruent to the initial unit square, so as to fill, without gaps and overlaps, the space immediately

surrounding it. This gives a bigger square the area of which is nine times bigger that the initial unit square. Make this bigger square the next unit square or “center square” for the construction and proceed as before. We consequently get a bigger square the area of which is eighty-one times bigger than the initial unit square and nine times bigger than the second unit square. Proceeding inductively, in each step magnifying by nine times the unit square of the previous step then constructing, without gaps and overlaps, eight surrounding congruent squares, it can be easily seen that the plane may be completely tiled. Note that the tiled area increases by a factor of nine in this procedure. The magnification need not be by nine only. Similar inductive processes may magnify the tiled area by a factor bigger than nine. Magnified induction for the equilateral triangle may be done in the same manner, i.e., by starting with a unit equilateral triangle or “center equilateral triangle”, building about it, and then using it as the next basic polygon. It uses the fact that any equilateral triangle may be divided into four smaller congruent triangles by using the three line segments determined by the midpoints of the three sides. Although magnified induction does not work for the regular hexagon one can see that such a tessellation can be constructed based on the tessellation of equilateral triangles since six non-overlapping equilateral triangles having a common vertex form a regular hexagon. Another possible method for tessellating the entire plane is what the author calls translated induction. The procedures are as follows. Start out with the basic unit square or “center square”. Then as in magnified induction, construct, about this “corner square”, eight squares congruent to this “corner square” so as to form, without gaps and overlaps, a bigger square the size of which is nine times the initial square. Now consider one of the eight newly constructed squares. Let this be the new “center square” and, as for the initial “center square”, add the remaining squares so as to form a square nine times bigger than the basic unit square. Do a similar construction for each of the remaining seven squares surrounding the initial basic unit square. Proceed inductively by similarly considering each square in the periphery of the current geometric configuration. By proceeding in this manner the plane can be tiled completely. Using translated induction, one can similarly tile the plane with equilateral triangles or regular hexagons.

Section 4. Discussion The proof given in Section 2 and the methods in Section 3 prove completely the following well-known result which was also previously stated as a proposition in [1]. Proposition 2. The only regular n-gons that can tile the plane completely are the equilateral triangle, the square, and the regular hexagon. Note that the interior angles of the equilateral triangle and the square (each measuring 60° and 90°, respectively) divide both 180° and 360° while the interior angles of a regular hexagon (each measuring 120°) divide 360° but not 180°. Observe that the equilateral triangle and the square, each of which has interior angles that divide 180°, are the only regular polygons that allow slide tessellations. Finally, a “slide version” of a tessellation, because of the way it is defined, can be obtained from the original tessellation by sliding. Hence, if one argues using this fact, the proof of Case II presented in Section 1 above may not be necessary for a complete proof.

References 1. Uro, Justine, “On Tessellating the Plane with Regular Convex Polygons,” Matimyas Matematika, vol. 15 (1992), pp. 4-8. 2. Bryant, S.J., Graham, G.E., and K.G. Wiley. Nonroutine Problems, [McGraw-Hill, Inc. 1965].