The Archimedean Tilings III- The Seeds of the Tilings

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The Archimedean Tilings III- The Seeds of the Tilings L.A. Romero 1 The Seed of an Archimdean Tiling A seed of an Archimedean tiling is a minimal group of tiles that can be translated in two directions to generate the tiling.

If you join the bottom orange edge to the top orange edge, you will generate a line of these seeds.

1 The Archimedean Tilings III- The Seeds of the Tilings L.A. Romero 1 The Seed of an Archimdean Tiling A seed of an Archimedean tiling is a minimal group of tiles that can be translated in two directions to generate the tiling. By minimal, we mean that there is no smaller set of tiles that can generate the tiling. Fig. 1 shows how to generate the snub squares (3, 3, 4, 3, 4) tiling from a seed. If you join the bottom orange edge to the top orange edge, you will generate a line of these seeds. If you then join the left green edge to the right green edge, you will generate the snub squares tiling. The number of each type of tile in the seed is the same for all seeds. However, the particular arrangement of the tiles can be changed to give a different seed. Fig. 2 shows an alternative seed for the snub squares (3, 3, 4, 3, 4) tiling. Figure 1: a) A seed for the snub squares tiling.. b) The snub squares (3, 3, 4, 3, 4) tiling Figure 2: An alternative seed for the snub squares (3, 3, 4, 3, 4) tiling. Other than this last example, in these notes, we will only give a single seed for each Archimedean tiling. Figs. 3, 4, 5, 6, 7, 8, and 9 show the seeds for the rest of the Archimedean tilings. 1

2 Figure 3: a) A seed for the truncated squares tiling.. b) The truncated squares (4, 8, 8) tiling Figure 4: a) A seed for the truncated hexagons tiling.. b) The truncated hexagons (3, 12, 12) tiling Figure 5: a) A seed for the tri-hex tiling.. b) The tri-hex (3, 6, 3, 6) tiling Figure 6: a) A seed for the snub hexagons tiling.. b) The snub hexagons (3, 3, 3, 3, 6) tiling 2

Figure 7: a) A seed for the rhombi tri-hex tiling.. b) The rhombi tri-hex (3, 4, 6, 4) tiling Figure 8: a) A seed for the great-hombi tri-hex tiling.

. b) The no-name (3, 3, 3, 4) tiling 2 The Relative Proportions of Tiles in an Archimedean Tiling In this section we will show a simple way of computing the relative proportions of each type of tile

If we build a finite (but very large) truncated squares tiling, the number of squares can be counted using number of squares (number of vertices) ( number of squares at a vertex)/4 (2.

3 Figure 7: a) A seed for the rhombi tri-hex tiling.. b) The rhombi tri-hex (3, 4, 6, 4) tiling Figure 8: a) A seed for the great-hombi tri-hex tiling.. b) The great-rhombi tri-hex (4, 6, 8) tiling Figure 9: a) A seed for the no-name tiling.. b) The no-name (3, 3, 3, 4) tiling 2 The Relative Proportions of Tiles in an Archimedean Tiling In this section we will show a simple way of computing the relative proportions of each type of tile in an Archimedean tiling. Since the tiling can be generated by translating the seed, this will also give us the relative proportions of the tiles in the seed. We will begin with a simple example. If we build a finite (but very large) truncated squares tiling, the number of squares can be counted using number of squares (number of vertices) ( number of squares at a vertex)/4 (2.1) This formula was arrived at by counting the tiles at every vertex. It is only approximate since it does not hold for vertices on the boundary of our tiling. However, as our tiling gets bigger, these boundary vertices will become less and less important. In this equation we need to divide by 4 since every square has four vertices, and hence will be counted 4 times if we do not divide by 4. 3

4 Similar reasoning gives number of octagons (number of vertices) ( number of octagons at a vertex)/8 (2.2) Using these formulas, we see that the number of octagons divided by the number of squares is approximatley equal to one. As the tiling gets larger (so that there are fewer boundary vertices), it will get arbitrarily close to one. Note that this is consistant with the fact that there are the same number of octagons as squares in the seed for the truncated squares tiling. This argument can be extended to the general case. Theorem 2.1. Suppose we have an Archimedean tiling with N different types of tiles. Let q k, k = 1, N be the number of each tile meeting at a vertex, and n k, k = 1, N be the number of edges on the kth type of tile. The relative proportions of the different types of tiles are (q 1 /n 1,...q N /n N ). This will give the relative proportions of the tiles in the seed. Example 1. What is the relative proportions of the tiles in the seed of the truncated squares (4, 8, 8) tiling? Answer We have (N squares, N octagons ) = k(1/4, 2/8) = k(1/4, 1/4). Thus there are the same number of octagons as squares in the seed. Example 2. What is the relative proportions of the tiles in the seed of the truncated hexagons (3, 12, 12) tiling? Answer We have (N dodecagons, N triangles ) = k(2/12, 1/3) = k(1/6, 1/3). Thus there are twice as many triangles in the seed as dodecagons. Example 3. What is the relative proportions of the tiles in the seed of the snub squares (3, 3, 4, 3, 4) tiling? Answer We have (N squares, N triangles ) = k(2/4, 3/3) = k(1/2, 1). Thus there are twice as many triangles in the seed as squares. Example 4. What is the relative proportions of the tiles in the seed of the rhombi-trihex (3, 4, 6, 4) tiling? Answer We have (N hexagons, N squares, N triangles ) = k(1/6, 2/4, 1/3) = (k/6)(1, 3, 2). Thus there are three times as many squares as hexagons, and two times as many triangles as hexagons. 3 The Average Number of Edges per Tile in an Archimedean Tiling In our notes on the topology of tiling we noted that for any tiling if p is the average number of edges per tile, and q is the average number of edges per vertex, then we must have 1 p + 1 q = 1 2 For a given Archimedean tiling, the number of edges at every vertex is the same. Hence it is very simple to compute q. Once we know the seed, it is also very simple to compute p. Theorem 3.1. In an Archimedean tiling, the average number of edges per face in the tiling is the same as the average number of edges per face in the seed. As an example. The average number of edges per face in the seed of the truncated squares (4, 8, 8) is p = (4 + 8)/2 = 6. Since the average number of edges per vertex is q = 3, we clearly have 1/p + 1/q = 1/2. Example 5. In the truncated hexagons (3, 12, 12) tiling the seed consists of a dodecagon and 2 triangles. The average number of edges per face is (12+3+3)/3 = 6. Since q = 3 in this tiling, we have 1/p+1/q = 1/2. Example 6. In the rhombi trihex (3, 4, 6, 4) tiling the seed consists of one hexagon, three squares and two triangles. The average number of edges per face is p = ( )/6 = 4. Since q = 4 in this tiling, we have 1/p + 1/q = 1/2. (3.1) 4

5 Example 7. In the snub squares (3, 3, 4, 3, 4) tiling the seed consists of four triangles and two squares. The average number of edges per face is p = ( )/6 = 10/3. Since q = 5 in this tiling, we have 1/p + 1/q = 1/2. 5

9.1. LEARN ABOUT the Math. How can you write the pattern rule using numbers and variables? Write a pattern rule using numbers and variables.

9.1. LEARN ABOUT the Math. How can you write the pattern rule using numbers and variables? Write a pattern rule using numbers and variables. 9.1 YOU WILL NEED coloured square tiles GOAL Write a pattern rule using numbers and variables. LEARN ABOUT the Math Ryan made this pattern using coloured tiles. relation a property that allows you to use