Tessellations. Irena Swanson Reed College, Portland, Oregon. MathPath, Lewis & Clark College, Portland, Oregon, 24 July 2018
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1 Tessellations Irena Swanson Reed College, Portland, Oregon MathPath, Lewis & Clark College, Portland, Oregon, 24 July 2018
2 What is a tessellation? A tiling or a tessellation of the plane is a covering of the plane with various (closed and countably many) shapes and with no overlaps and no gaps (other than overlaps on the boundaries of the shapes).
3 What is a tessellation? A tiling or a tessellation of the plane is a covering of the plane with various (closed and countably many) shapes and with no overlaps and no gaps (other than overlaps on the boundaries of the shapes). Some possible tiles: and and
4 M. C. Escher
5 M. C. Escher
6 M. C. Escher
7 For today allow only regular polygons for tiles.
8 For today allow only regular polygons for tiles. Some possible tessellations with regular polygons:
9 For today allow only regular polygons for tiles. Some possible tessellations with regular polygons: For today allow only edge-to-edge tessellations (only the leftmost example).
10
11 Stack these rows, matching vertices, but vary the numbers of types of strips. You get uncountably many tessellations.
12 Stack these rows, matching vertices, but vary the numbers of types of strips. You get uncountably many tessellations. Or tessellate with identical rows of hexagons and triangles in uncountably many ways by stacking them in two different ways:
13 Type of a vertex: all
14 Type of a vertex: all For now we restrict attention to edge-to-edge tessellations by regular polygons in which each vertex has the same type.
15 Type of a vertex: all For now we restrict attention to edge-to-edge tessellations by regular polygons in which each vertex has the same type. Such tessellations are called semiregular tessellations, or also homogeneous tessellations, Archimedean tessellations, 1-uniform tessellations.
16 Determining all semiregular tessellations
17 Determining all semiregular tessellations Some examples: (1) Type (all squares) (2) Type (all triangles) (3) Type (all hexagons) (4) Type
18 Determining all semiregular tessellations Some examples: (1) Type (all squares) (2) Type (all triangles) (3) Type (all hexagons) (4) Type (5) Type
19 Determining all semiregular tessellations Some examples: (1) Type (all squares) (2) Type (all triangles) (3) Type (all hexagons) (4) Type (5) Type There are others. How do we find them all?
20 Theorem. A regular polygon with n sides (n-gon) has interior angles equal n in degrees.
21 Theorem. A regular polygon with n sides (n-gon) has interior angles equal n in degrees. Here is an n-gon:
22 Theorem. A regular polygon with n sides (n-gon) has interior angles equal n in degrees. Here is an n-gon: The sum of all the angles in all n triangles is n 180,
23 Theorem. A regular polygon with n sides (n-gon) has interior angles equal n in degrees. Here is an n-gon: The sum of all the angles in all n triangles is n 180, the sum of all the angles at the central vertex is 360,
24 Theorem. A regular polygon with n sides (n-gon) has interior angles equal n in degrees. Here is an n-gon: The sum of all the angles in all n triangles is n 180, the sum of all the angles at the central vertex is 360, so the sum of all interior angles in an n-gon is (n 2)180.
25 Theorem. A regular polygon with n sides (n-gon) has interior angles equal n in degrees. Here is an n-gon: The sum of all the angles in all n triangles is n 180, the sum of all the angles at the central vertex is 360, so the sum of all interior angles in an n-gon is (n 2)180. Thus an interior angle in a regular n-gon measures n 2 n 180.
26 From the previous page: An interior angle in a regular n-gon measures n 2 n 180.
27 From the previous page: An interior angle in a regular n-gon measures n 2 n 180. In a semiregular tessellation in which each vertex has type n 1.n n r, the angles of these gons have to add up to 360.
28 From the previous page: An interior angle in a regular n-gon measures n 2 n 180. In a semiregular tessellation in which each vertex has type n 1.n n r, the angles of these gons have to add up to 360. Thus we get: Theorem. In order for a regular n 1 -gon, n 2 -gon,..., and an n r -gon to meet at a vertex without overlaps and without gaps, it is necessary and sufficient that n 1 2 n n 2 2 n n r 2 n r 180 = 360.
29 Theorem. In order for a regular n 1 -gon, n 2 -gon,..., and an n r -gon to meet at a vertex without overlaps and without gaps, it is necessary and sufficient that n 1 2 n 1 + n 2 2 n n r 2 n r = 2.
30 Theorem. In order for a regular n 1 -gon, n 2 -gon,..., and an n r -gon to meet at a vertex without overlaps and without gaps, it is necessary and sufficient that Thus n 1 2 n 1 + n 2 2 n n r 2 n r = 2. r 2 = 2 n n n r
31 Theorem. In order for a regular n 1 -gon, n 2 -gon,..., and an n r -gon to meet at a vertex without overlaps and without gaps, it is necessary and sufficient that Thus n 1 2 n 1 + n 2 2 n n r 2 n r = 2. r 2 = 2 n n n r
32 Theorem. In order for a regular n 1 -gon, n 2 -gon,..., and an n r -gon to meet at a vertex without overlaps and without gaps, it is necessary and sufficient that Thus n 1 2 n 1 + n 2 2 n n r 2 n r = 2. r 2 = 2 n n n r = r 2 3
33 Theorem. In order for a regular n 1 -gon, n 2 -gon,..., and an n r -gon to meet at a vertex without overlaps and without gaps, it is necessary and sufficient that Thus n 1 2 n 1 + n 2 2 n n r 2 n r = 2. r 2 = 2 n n n r = r 2 3 so that which means that r 6. r 3 = r r 2 3 2,
34 So, we have to find all integer solutions n 1, n 2,..., n r 3 of r 2 = 2 n n n r for r = 3, r = 4, r = 5 and r = 6.
35 So, we have to find all integer solutions n 1, n 2,..., n r 3 of r 2 = 2 n n n r for r = 3, r = 4, r = 5 and r = 6. This is now number theory, via case analysis. (These equations are Diophantine.)
36 So, we have to find all integer solutions n 1, n 2,..., n r 3 of r 2 = 2 n n n r for r = 3, r = 4, r = 5 and r = 6. This is now number theory, via case analysis. (These equations are Diophantine.) For example, let r = 3.
37 So, we have to find all integer solutions n 1, n 2,..., n r 3 of r 2 = 2 n n n r for r = 3, r = 4, r = 5 and r = 6. This is now number theory, via case analysis. (These equations are Diophantine.) For example, let r = 3. Then we need to find n 1, n 2 and n 3 with 3 2 = 2 n n n 3, i.e., with 1 2 = 1 n n n 3.
38 So, we have to find all integer solutions n 1, n 2,..., n r 3 of r 2 = 2 n n n r for r = 3, r = 4, r = 5 and r = 6. This is now number theory, via case analysis. (These equations are Diophantine.) For example, let r = 3. Then we need to find n 1, n 2 and n 3 with 3 2 = 2 n n n 3, i.e., with 1 2 = 1 n n n 3. If n 1 n 2 n 3, then n 1 6.
39 So, we have to find all integer solutions n 1, n 2,..., n r 3 of r 2 = 2 n n n r for r = 3, r = 4, r = 5 and r = 6. This is now number theory, via case analysis. (These equations are Diophantine.) For example, let r = 3. Then we need to find n 1, n 2 and n 3 with 3 2 = 2 n n n 3, i.e., with 1 2 = 1 n n n 3. If n 1 n 2 n 3, then n 1 6. If n 1 = 3, then we need to find n 2 and n 3 with 1 6 = 1 n n 3.
40 Solving 1 6 = 1 n n 3 (or n 3 = n 2 6 ):
41 Solving 1 6 = 1 n n 3 (or n 3 = n 2 6 ): All solutions (with r = 3, n 1 = 3 n 2 n 3 ): n 2 = 7, n 3 = 42, n 2 = 8, n 3 = 24, n 2 = 9, n 3 = 18, n 2 = 10, n 3 = 15, n 2 = 12, n 3 = 12.
42 Solving 1 6 = 1 n n 3 (or n 3 = n 2 6 ): All solutions (with r = 3, n 1 = 3 n 2 n 3 ): n 2 = 7, n 3 = 42, n 2 = 8, n 3 = 24, n 2 = 9, n 3 = 18, n 2 = 10, n 3 = 15, n 2 = 12, n 3 = 12. Still need all solutions to r 2 = 2 n n n r for r = 3, r = 4, r = 5 and r = 6.
43 The complete list of integer solutions n 1,..., n r 3 for the solutions of if we ignore the order of the n i, is: n 1 2 n 1 + n 2 2 n n r 2 n r = 2, (1) 3, 3, 3, 3, 3, 3 (2) 3, 3, 3, 3, 6 (3) 3, 3, 3, 4, 4 (4) 3, 3, 4, 12 (5) 3, 3, 6, 6 (6) 3, 4, 4, 6 (7) 3, 7, 42 (8) 3, 8, 24 (9) 3, 9, 18 (10) 3, 10, 15 (11) 3, 12, 12 (12) 4, 4, 4, 4 (13) 4, 5, 20 (14) 4, 6, 12 (15) 4, 8, 8 (16) 5, 5, 10 (17) 6, 6, 6
44 The complete list of integer solutions n 1,..., n r 3 for the solutions of if we ignore the order of the n i, is: n 1 2 n 1 + n 2 2 n n r 2 n r = 2, (1) 3, 3, 3, 3, 3, 3 (2) 3, 3, 3, 3, 6 (3) 3, 3, 3, 4, 4 (4) 3, 3, 4, 12 (5) 3, 3, 6, 6 (6) 3, 4, 4, 6 (7) 3, 7, 42 (8) 3, 8, 24 (9) 3, 9, 18 (10) 3, 10, 15 (11) 3, 12, 12 (12) 4, 4, 4, 4 (13) 4, 5, 20 (14) 4, 6, 12 (15) 4, 8, 8 (16) 5, 5, 10 (17) 6, 6, 6 Warning: not all of these are realizable semiregular tessellations;
45 The complete list of integer solutions n 1,..., n r 3 for the solutions of if we ignore the order of the n i, is: n 1 2 n 1 + n 2 2 n n r 2 n r = 2, (1) 3, 3, 3, 3, 3, 3 (2) 3, 3, 3, 3, 6 (3) 3, 3, 3, 4, 4 (4) 3, 3, 4, 12 (5) 3, 3, 6, 6 (6) 3, 4, 4, 6 (7) 3, 7, 42 (8) 3, 8, 24 (9) 3, 9, 18 (10) 3, 10, 15 (11) 3, 12, 12 (12) 4, 4, 4, 4 (13) 4, 5, 20 (14) 4, 6, 12 (15) 4, 8, 8 (16) 5, 5, 10 (17) 6, 6, 6 Warning: not all of these are realizable semiregular tessellations; solution 3, 3, 3, 4, 4 yields two semiregular tessellations.
46 Eliminate some options The vertex configuration 3, 3, 4, 12 on the list could yield vertex types or Neither is possible. I show how to eliminate
47 Eliminate some options The vertex configuration 3, 3, 4, 12 on the list could yield vertex types or Neither is possible. I show how to eliminate
48 Eliminate some options The vertex configuration 3, 3, 4, 12 on the list could yield vertex types or Neither is possible. I show how to eliminate Forced to continue with a triangle: but a triple triangle vertex is not allowed for 3, 3, 4, 12.
49 Theorem. There are 11 semi-regular tessellations up to translations, rotations and reflections: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) 6.6.6
50 Theorem. There are 11 semi-regular tessellations up to translations, rotations and reflections: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) Compare to all solutions from before: (1) 3, 3, 3, 3, 3, 3 (2) 3, 3, 3, 3, 6 (3) 3, 3, 3, 4, 4 (4) 3, 3, 4, 12 (5) 3, 3, 6, 6 (6) 3, 4, 4, 6 (7) 3, 7, 42 (8) 3, 8, 24 (9) 3, 9, 18 (10) 3, 10, 15 (11) 3, 12, 12 (12) 4, 4, 4, 4 (13) 4, 5, 20 (14) 4, 6, 12 (15) 4, 8, 8 (16) 5, 5, 10 (17) 6, 6, 6
51
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