CHAPTER 12

TESSELLATIONS

12.1 Tilings

A tiling or tessellation of the plane is a division of the plane into regions

called tiles, in such a manner that:

In other words, the plane is completely covered by nonoverlapping tiles.

There is no requirement that the tiles be related in any way, but our interest is primarily in tilings where there are only a finite number of differently shaped tiles. A tessellation is of order-k, or k-hedral, if there is a finite set of k incongruent tiles S such that:

The members of S are called prototiles, and we say that S tiles the plane. The tiling in the figure below has a set of three prototiles, so it is an order-3 tiling.

12.2 Monohedral Tilings

The most natural question is “Which polygons are monohedral prototiles?” The words monohedral, dihedral, and trihedral are commonly used as synonyms for 1-hedral, 2-hedral, and 3-hedral, respectively.

It is not difficult to see that every parallelogram tiles the plane, as in (a) below.

As a consequence, every triangle tiles the plane because two copies of the triangle can be combined to form a parallelogram, as in (b) above.

Every quadrilateral tiles the plane. The tiling can be obtained by successively rotating the quadrilateral around the midpoints of the sides, as in the figure below.

Tile #1 is obtained by applying H_A to the shaded tile, and tile #2 is obtained by applying H_B to tile #1. Continuing in this way, we can tile the entire horizontal strip containing tiles #1 and #2. The strip below this one can be obtained by similar rotations—for example, tile #3 may be obtained by applying H_C to the shaded tile. Alternatively, we can apply T_XY to the entire strip containing #1 and #2.

The regular pentagon will not tile the plane. One way to see this is to examine what happens when you try to tile around a vertex.

The figures above show what happens when you try to tile around vertex X of the shaded pentagon. You can either try a vertex-to-vertex tiling, as in (a), or you can try an edge-to-vertex tiling, as in (b). In (a) there is an angular gap of 36°, and in (b) there is a 72° gap, neither of which can be covered without overlapping tiles.

Some hexagons tile the plane, but not all do. Everyone has seen the tiling in (a) below.

The hexagon in (b) above will not tile the plane, and, as before, this can be verified by trying to tile around the vertex of x°.

Although the regular pentagon does not tile the plane, there are some pentagons that do, and the figure below shows two of them.

The tiling in (a) is obtained by splitting the regular hexagonal tiling.

The tiling in (b), which also uses hexagons, is the beautiful “Cairo” tiling.

Convex Hexagons that Tile

A polygon is convex if all of its diagonals are interior to the polygon.

As shown earlier, some convex pentagons and some convex hexagons tile the plane. The situation for convex hexagons is completely understood—a convex polygon ABCDEF will tile the plane if and only if it satisfies one of the following three criteria:

The figures below give an example of each type.

However, it is still not known which convex pentagons tile the plane. At the time of writing this text, we know that 15 different types of convex pentagons tile the plane. This problem has a fascinating history, and you can find much more information about the stale of affairs of this problem on the web or in the following articles:

“Tiling with Convex Polygons,” Chapter 13 of Time Travels and Other Mathematical Bewilderments (Martin Gardner; W. H. Freeman and Company, New York, 1988).

“In Praise of Amateurs.” Doris Schattschneider, in The Mathematical Gardner (David A. Klarner, Editor. Prindle, Weber, and Schmidt, Boston, 1981).

12.3 Tiling with Regular Polygons

In this section, we will investigate tessellations or tilings with regular polygons whose sides are of unit length and such that two neighboring tiles share a complete edge. Such tessellations are said to be edge-to-edge tilings. The diagram below shows three familiar examples of monohedral tessellations of this type.

Recall that in order to form a tessellation, the tiles must be able to completely surround a vertex without overlapping.

The sequence of regular polygonal tiles that surround a point generates a vertex sequence of the point which gives the number of sides of each tile.

For example, in the first tessellation above, the vertex sequence of each vertex is (3,3,3,3,3,3). In the second and third tessellations, the vertex sequences are (4,4,4,4) and (6,6,6), respectively.

We will show that there are other combinations of regular polygonal tiles that can surround a point. In order to do this, we need to find a set of angles whose total measure is 360° and such that the measure of each angle is the measure of a vertex angle of an n-sided regular polygon, that is,

where n is an integer greater than or equal to 3.

Combinations of Three Polygons

We first consider combinations with three polygons. Let the vertex sequence be (a, b, c) with a ≤ b ≤ c. Then

which simplifies to

Case (i). a = 3.

In this case, the equation above becomes

Solving for c in terms of b, we have

so that

and this implies that b > 6.

For b = 7, we have c = 42; for b = 8, we have c = 24; for b = 9, c = 18; for b = 10, c = 15; for b = 11, c is not an integer; and for b = 12, c = 12.

Since b ≤ c, the only solutions in this case are (3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), and (3, 12, 12).

Note. Vertex sequences that are not monohedral can have equivalent forms. For instance, the vertex sequence (3,7,42) has five other equivalent forms, namely,

They are listed in lexicographical order, that is, a vertex sequence with a smaller first term comes before one with a larger first term. Among those with equal first terms, a vertex sequence with a smaller second term comes before one with a larger second term, and so on.

Case (ii). a = 4.

In this case, the equation

becomes

Solving for c in terms of b, we have

so that

and this implies that b > 4.

For b = 5, c = 20; for b = 6, c = 12; for b = 7, c is not an integer; and for b = 8, c = 8.

Since b ≤ c, the only solutions in this case are

Case (iii). a = 5.

In this case, the equation

becomes

Solving for c in terms of b, we have

so that

and we have b ≥ a = 5.

For b = 5, c = 10; for b = 6, 7, c is not an integer; and for b = 8, we have b > c. Thus, the only solution in this case is (5,5,10).

Case (iv). a = 6.

In this case, if c ≥ 7, then

which is a contradiction. Hence,

and the only solution in this case is (6,6,6).

Case (v). a ≥ 7.

In this case,

which is a contradiction. Hence, there are no further solutions.

Note. In the case (3,7,42), neither the regular 7-gon nor the regular 42-gon has angles with integral measures. Thus, this combination is not easy to discover just by inspection.

Combinations of Four Polygons

Next we consider combinations with four polygons. Let the vertex sequence be a permutation of (a, b, c, d) with a ≤ b ≤ c ≤ d. As before, we have

Case (i). a = 3.

In this case, the equation above becomes

If b = 3. we have

and solving for d in terms of c, we have

so that

and therefore c > 3.

For c = 4, d = 12; for c = 5, d is not an integer; and for c = 6, d = 6.

Since c ≤ d, the only solutions for b = 3 are (3,3,4,12) and (3,3,6,6).

If b = 4, we have

and solving for d in terms of c, we have

so that

and therefore c ≥ b = 4.

For c = 4, d = 6; for c = 5, we already have c > d. Thus, the only solution when b = 4 is (3,4,4,6).

For b ≥ 5,

which is a contradiction. Hence, there are no further solutions in this case.

Case (ii). a = 4.

In this case, for d ≥ 5,

which is a contradiction. Hence a = b = c = d = 4, and the only solution in this case is (4,4,4,4).

Case (iii). a ≥ 5.

In this case,

which is a contradiction. Hence there are no further solutions.

Combinations of Five Polygons

Now we consider combinations with five polygons. Let the vertex sequence be a permutation of (a, b, c, d, e) where a ≤ b ≤ c ≤ d ≤ e. As before, we have

For c ≥ 4,

which is a contradiction.

Hence, a = b = c = 3, so that

As before, we have

so that the only solutions in this case are (3,3,3,3,6) and (3,3,3,4,4).

Combinations of Six Polygons

Finally, we consider combinations with six polygons. Let the vertex sequence be a permutation of (a, b, c, d, e, f) where a ≤ b ≤ c ≤ d ≤ e ≤ f. As before, we have

For f ≥ 4,

which is a contradiction. Hence, a = b = c = d = e = f = 3, and the only solution in this case is (3,3,3,3,3,3).

Note. We cannot surround a point with seven or more regular polygons since the smallest of the angles at this point is at least 60°, and the sum of these angles will exceed 6 × 60° = 360°.

Several of our solutions give rise to more than one vertex sequence. The combination (3,3,4,12) may be permuted as (3,4,3,12). The combination (3,3,6,6) maybe permuted as (3,6,3,6). The combination (3,4,4,6) may be permuted as (3,4,6,4). Finally, the combination (3,3,3,4,4) may be permuted as (3,3,4,3,4). There are left-handed and right-handed versions of (3,3,3,3,6), but they are not considered to be different. This brings the total number of possible vertex sequences to 21.

12.4 Platonic and Archimedean Tilings

Now we consider tilings that are named after the Greek philosophers Plato and Archimedes.

For each vertex sequence, we wish to know if we can tile the entire plane with regular polygons such that every vertex has this vertex sequence. In this case, such a tessellation is said to be semiregular.

Moreover, if all the terms in the vertex sequences are identical, the tessellation is said to be regular. The regular tessellations of the plane are called Platonic tilings.

Semiregular tessellations that are not regular are called Archimedean tilings.

We divide the 21 possible vertex sequences into three groups.

• Group I.

These vertex sequences are all of the form (a, b, c) where a is odd and b ≠ c if we read (5,4,20) for (4,5,20).

If there is a tessellation in which every vertex has this vertex sequence, we must be able to surround the a-sided polygon. Thus, its neighbours must be the b-sided polygon and the c-sided polygon, alternately. However, this is impossible since a is odd. The figure below illustrates the case (5,5,10). Moreover, no combination of regular polygons can fill the void.

• Group II.

Each of these vertex sequences contains a triangular tile. We can surround two of its vertices properly, but the third vertex, indicated with a black dot in the diagram on the following page, requires a different vertex sequence.

• Group III.

Here we have no local problem. It turns out that each of these vertex sequences leads to a semiregular tessellation. There are exactly three Platonic tessellations, namely,

which we saw at the beginning of Section 12.3.

The other eight are the only Archimedean tilings, and they will be discussed a later. However, we still have to prove that we have no global problem with any of these 11 vertex sequences. We shall use a direct approach and construct each of the 11 tessellations.

(a) The (3,3,3,3,3,3) and the (3,6,3,6) tessellations may be constructed with three infinite families of evenly spaced parallel lines forming 60° angles across the families, as shown in the diagram below.

(b) The (4,4,4,4) tessellation may be constructed with two infinite familes of evenly spaced parallel lines forming 90° angles across families, as shown in the figure below on the left. This tessellation and the (3,3,3,3,3,3) and (3,6,3,6) tessellations are the three basic tessellations.

(c) The (3,3,3,4,4) tessellation is obtained by taking alternate strips from the basic (3,3,3,3,3,3) and (4,4,4,4,4) tessellations, as in the figure above on the right.

The remaining tessellations are obtained from others by the cut-and-merge method.

(d) From the basic (3,3,3,3,3,3) tessellation, we can merge a set of six equilateral triangles into a regular hexagon, as in the figure below on the left.

(e) The (6,6,6) and the (3,3,3,3,6) tessellations may be obtained from the basic (3,3,3,3,3,3) tessellations by merging various sets of six equilateral triangles, as shown in the figure above in the middle and on the right. No cutting is required in either case.

(f) The (3,4,6,4) tessellation can also be obtained from the basic (3,3,3,3,3,3) tessellation. As shown in the figure below on the left, we cut each triangular tile into seven pieces consisting of an equilateral triangle, three congruent half-squares, and three congruent kites with angles 120°, 90°, 60°, and 90°.

Let the edge length of the triangular tile be 1 and the length of the side of the equilateral triangle be x. Then the short sides of the kite have length x/2 and the long sides √3x/2. Since

we have

When we merge the kites and half-squares across six triangular tiles, we obtain the (3,4,6,4) tessellations, as shown in the figure below on the right.

(g) The (4,6,12) tessellation can now be obtained from the (3,4,6,4) tessellation without cutting. Each dodecagon in the new tessellation is obtained by merging one regular hexagon, six squares and six equilateral triangles in the old tessellation, as shown in the figure on the following page.

(h) The (3,12,12) tessellation may be constructed from the (6,6,6) tessellation. We cut each hexagonal tile into seven pieces consisting of a regular dodecagon and six congruent isosceles triangles with vertical angles 120°, as in the figure on the following page on the left. Let the edge length of the hexagonal tile be 1 and the length of the base of the triangles be x. The equal sides of the triangles have length x/√3, and since

we have

When we merge the triangles across three hexagonal tiles, we obtain the (3,12,12) tessellation, as in the figure on the following page on the right.

(i) Next, we construct the (4,8,8) tessellation from the basic (4,4,4,4) tessellation. We cut each square tile into five pieces consisting of a regular octagon and four congruent right isosceles triangles, as in the figure below on the left. Let the edge length of the square tile be 1 and the length of the hypotenuse of the triangles be x. Then the legs of the triangles have length x/√2. Since

we have

When we merge the triangles across four square tiles, we obtain the (4,8,8) tessellation, as in the figure below on the right.

(j) The last tessellation, (3,3,4,3,4), is the most difficult to construct. It is obtained from the basic (4,4,4,4) tessellation with an intermediate step. We first modify the square tile as in the figure on the following page on the left. We cut out two isosceles triangles with vertical angles 150°, based on two opposite sides of the square, and attach them to the other two sides, This modified tile can also tile the plane, as in the figure on the following page on the right.

We now cut each modified tile into six pieces consisting of two congruent equilateral triangles and four congruent right isosceles triangles, as in the figure below on the left,

When we merge the right isosceles triangles across four modified tiles, we obtain the (3,3,4,3,4) tessellations, as in the figure below on the right.

The eight Archimedean tilings are graphically illustrated on page 222 in the book Sphere Packing, Lewis Carroll and Reversi by Martin Gardner, published in 2009 by the Mathematical Association of America, Washington, DC.

12.5 Problems

1. There are 12 ways in which five unit squares can be joined edge to edge. The resulting figures are called pentominoes, as shown in the figure on the following page.

They are given letter names, F, I, L, N, P, T, U, V, W, X, Y, and Z, respectively. Identify those that can tile a rectangle.

2. The figure below shows the tiling of a bent strip by a figure consisting of four unit squares joined edge to edge.

Obviously, a pentomino that can tile a rectangle can also tile some bent strip. Identify those that can tile a bent strip but not a rectangle.

3. Prove that a pentomino that can tile a bent strip can also tile some infinite strip.

4. Identify those pentominoes that can tile an infinite strip but cannot tile a bent strip.

5. Obviously, a pentomino that can tile an infinite strip can also tile the plane. Identify those that can tile the plane but cannot tile an infinite strip.

6. For each of the I-pentomino, L-pentomino, N-pentomino, U-pentomino, V-pentomino, and W-pentomino, dissect it into four pieces and reassemble them into a square.

7.   (a) Show that the figure below can tile the plane.

(b) For each of the P-pentomino, Y-pentomino, and Z-pentomino, dissect it into three pieces and reassemble them into a square.

8.   (a) Show that the figure below can tile the plane.

(b) For each of the F-pentomino, T-pentomino, and X-pentomino, dissect it into four pieces and reassemble them into a square.

9. Dissect the figure below into three pieces and reassemble them into a square.

10. Dissect the figure below into three pieces and reassemble them into an equilateral triangle.

11.  (a) Prove that the sum of measures of the exterior angles of a regular n-gon is 360° for any n ≥ 3.

(b) Use part (a) to prove that the sum of the measures of the central angles of a regular n-gon is given by (n − 2)180°.

12. For which values of n is the measure of the central angle of an n-sided regular polygon a positive integer?

13. Find three ways of obtaining the basic (3,6,3,6) tessellation from other tessellations.

14. Find another way of obtaining the (6,6,6) tessellation from the basic (3,3,3,3,3,3) tessellation.

15. Find a way of obtaining the (3,4,6,4) tessellation from the (6,6,6) tessellation.

16. Find a tessellation that has exactly two kinds of vertex sequences, one of which is (3,3,4,12).

17. Find a tessellation that has exactly two kinds of vertex sequences, one of which is (3,4,3,12).

18. Find three tessellations that have exactly two kinds of vertex sequences, one of which is (3,3,6,6).

19. Find three tessellations that have exactly two kinds of vertex sequences, one of which is (3,4,4,6).