Tessellations with Regular Polygons

1. Tessellations with Regular Polygons Lecturer : Professor Andy Liu Chairman, Academic Committee of IMC Professor, Department of Mathematical and Statistical Sciences University of Alberta, Canada.

2. Section 1. Regular Polygons

3. Section 1. Regular Polygons Let n be an integer greater than or equal to 3. A regular n-gon is a polygon with n sides such that all sides have the same length and all angles have the same measure. The regular 3-gon is called the equilateral triangle. The sum of the measures of its three angles is 180°. Hence the measure of each of its angles is 180° ÷ 3 = 60 °.

4. Section 1. Regular Polygons The regular 4-gon is called the square. It may be dissected into two triangles by either of its diagonals. It follows that the sum of the measures of its four angles is 2 × 180°= 360°. Hence the measure of each of its angles is 360° ÷ 4 = 90°.

5. Section 1. Regular Polygons In general, the regular n-gon may be dissected into n－2 triangles by its diagonals from any of its vertices. It follows that the sum of the measures of its n angles is (n － 2)180°. Hence the measure of each of its angles is

6. Section 1. Regular Polygons For which values of n is the measure of the angles an integer? Let this integer be k. Then so that n(180 － k) °=360°. It follows that n must be a divisor of 360. The chart below gives the values of these measures for each divisor of 360 other than 1 and 2.

7. Section 1. Regular Polygons

8. Section 1. Regular Polygons Note that 360° ÷ 60°= 6, 360°÷ 90°= 4 and 360°÷ 120°= 3. These are the only cases with integral quotients. Thus we can only surround a point with the same kind of regular polygons, without overlap, by using 6 equilateral triangles, 4 squares or 3 regular hexagons, as shown in Figure 1.

9. Section 1. Regular Polygons We shall describe a vertex by the numbers of sides of the polygons surrounding it. Thus the three vertices above may be described as (3, 3, 3, 3, 3, 3), (4, 4, 4, 4) and (6, 6, 6). Such a sequence is called a vertex sequence.

10. Section 1. Regular Polygons Are there other combinations of polygons which can surround a point? We seek a set of angles whose total measure is 360°. We can find some of them by inspecting the values in the chart above, but let us use a more system approach.

11. Section 1. Regular Polygons We first consider combinations with three polygons. Let the vertex sequence be (a, b, c), a≦b≦c. Then which simplifies to

12. Section 1. Regular Polygons Suppose a=3. Solving for c in terms of b, we have so that

13. Section 1. Regular Polygons Hence b>6. For b=7, c=42. For b=8, c=24. For b=9, c=18. For b=10, c=15. For b=11, c is not an integer. 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).

14. Section 1. Regular Polygons Suppose a=4. Solving for c in terms of b, we have so that

15. Section 1. Regular Polygons Hence b>4. For b=5, c=20. For b=6, c=12. For b=7, c is not an integer. For b=8, c=8. Since b≦c, the only solutions in this case are (4, 5, 20), (4, 6, 12) and (4, 8, 8).

16. Section 1. Regular Polygons Suppose a=5. Solving for c in terms of b, we have so that

17. Section 1. Regular Polygons Recall that b≧a=5. For b=5, c=10. For b=6 or 7, c is not an integer. For b=8, we already have b>c. Thus the only solution in this case is (5, 5, 10).

18. Section 1. Regular Polygons Suppose a=6. Recall that c≧7, which is a contradiciton. Hence a=b=c=6 so that the only solution in this case is (6, 6, 6).

19. Section 1. Regular Polygons For a≧7, a contradiction. Hence there are no further solutions. Note that 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.

20. Section 1. Regular Polygons We next consider combinations with four polygons. Let the vertex sequence be a permutation of (a, b , c, d), a≦b≦c≦d. As before, we have

21. Section 1. Regular Polygons Suppose a=3. If b=3, we have Solving for d in terms of c, we have so that

22. Section 1. Regular Polygons Hence c>3. For c=4, d=12. For c=5, d is not an integer. For c=6, d=6. Since c≦d, the only solutions in this subcase are (3, 3, 4, 12) and (3, 3, 6, 6).

23. Section 1. Regular Polygons Suppose a=3. If b=4, we have Solving for d in terms of c, we have so that

24. Section 1. Regular Polygons Recall that c≧b=4. For c=4, d=6. For c=5, we already have c>d. Thus the only solution in this subcase is (3, 4, 4, 6).

25. Section 1. Regular Polygons Suppose a=3. If b≧5, , which is a contradiction. Hence there are no further solutions in this case

26. Section 1. Regular Polygons Suppose a=4. For d≧5, we have which is a contradiction. Hence a=b=c=d=4, so that the only solution in this case is (4,4,4,4).

27. Section 1. Regular Polygons For a≧5, which is a contradiction. Hence there are no further solutions.

28. Section 1. Regular Polygons Now we consider combinations with five polygons. Let the vertex sequence be a permutation of (a, b , c, d, e), a≦b≦c≦d≦e. As before, we have

29. Section 1. Regular Polygons For c≧4, which is a contradiction. Hence a=b=c=3 so that As before, we have (d, e)=(3, 6) or (4, 4), so that the only solutions in this case are (3, 3, 3, 3, 6) and (3, 3, 3, 4, 4).

30. Section 1. Regular Polygons Finally, we consider combinations with six polygons. Let the vertex sequence be a permutation of (a, b , c, d, e, f ), a≦b≦c≦d≦e≦f. As before, we have

31. Section 1. Regular Polygons For f ≧4, which is a contradiction. Hence a=b=c=d=e=f =3, so that the only solution in this case is (3, 3, 3, 3, 3, 3).

32. Section 1. Regular Polygons We cannot surround a point with seven or more 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°.

33. Section 1. Regular Polygons 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) may be 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).

34. Section 1. Regular Polygons There is a left-handed version and a right-handed version of (3, 3, 3, 3, 6), but they are not considered to be different. This brings the total number of vertex sequences to 21.

35. Section 2. Platonic and Archimedean Tilings

36. Section 2. Platonic and Archimedean Tilings 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. Such a tessellation is said to be semi-regular. If moreover all terms in the vertex sequences are identical, it is said to be regularand are called Platonictilings. Semi-regular tessellations which are not regular are called Archimedeantilings. They are named after two Greek philosophers.

37. Section 2. Platonic and Archimedean Tilings We consider the 21 possible vertex sequences in three groups. Group I：(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (4, 5, 20) and (5, 5, 10). Group II ：(3, 3, 4, 12), (3, 3, 6, 6), (3, 4, 3, 12), (3, 4, 4, 6). Group III ：(3, 12, 12), (4, 6, 12), (4, 8, 8), (6, 6, 6), (3, 4, 6, 4), (3, 6, 3, 6), (4, 4, 4, 4), (3, 3, 3, 3, 6), (3, 3, 3, 4, 4), (3, 3, 4, 3, 4), (3, 3, 3, 3, 3, 3).

38. Section 2. Platonic and Archimedean Tilings Group I：(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (4, 5, 20) and (5, 5, 10). These vertex sequences are all of the form (a, b, c) where a is odd and b≠c. For (4, 5, 20), read (5, 4, 20).

39. ？ (5, 5, 10) Section 2. Platonic and Archimedean Tilings If there is a tessellation in which every vertex has this vertex sequence, we must be able to surround the a-sided polygon. Its neighbors must be the b-sided and the c-sided polygon alternately. However, this is impossible since a is odd.

40. Section 2. Platonic and Archimedean Tilings Group II ：(3, 3, 4, 12), (3, 3, 6, 6), (3, 4, 3, 12), (3, 4, 4, 6). Here the problem is not local but global. We can indeed surround the largest polygon so that each of its vertices has the same vertex sequence. However, at least one other vertex cannot possibly have that vertex sequence.

41. (3, 4, 4, 6) (3, 3, 6, 6) (3, 3, 4, 12) (3, 4, 3, 12) Section 2. Platonic and Archimedean Tilings

42. Section 2. Platonic and Archimedean Tilings Group III ：(3, 12, 12), (4, 6, 12), (4, 8, 8), (6, 6, 6), (3, 4, 6, 4), (3, 6, 3, 6), (4, 4, 4, 4), (3, 3, 3, 3, 6), (3, 3, 3, 4, 4), (3, 3, 4, 3, 4), (3, 3, 3, 3, 3, 3). Here we have neither local problem nor global problem, so that there are exactly three Platonic tilings (3,3,3,3,3,3), (4,4,4,4) and (6,6,6). The other eight are the only Archimedean tilings.

43. Section 2. Platonic and Archimedean Tilings It is easy to verify that we have no local problem here, but it is not at all easy to prove that we have no global problem either. We shall use a direct approach by constructing each of the eleven tessellations.

44. Section 2. Platonic and Archimedean Tilings 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 families. (3, 3, 3, 3, 3, 3) (3, 6, 3, 6)

45. Section 2. Platonic and Archimedean Tilings The (4, 4, 4, 4) tessellation may be constructed with two infinite familes of evenly spaced parallel lines forming 90° angles across families. (4, 4, 4, 4) This tessellation and the (3, 3, 3, 3, 3, 3) and (3, 6, 3, 6) tessellations are the threebasictessellations.

46. Section 2. Platonic and Archimedean Tilings 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) tessellations. (3, 3, 3, 4, 4)

47. Section 2. Platonic and Archimedean Tilings The remaining tessellations are obtained from others by the cut and merge method. The basic (3,3,3,3,3,3) tessellation with a set of six equilateral triangles merged into a regular hexagon.

48. Section 2. Platonic and Archimedean Tilings The (6,6,6) and the (3,3,3,3,6) tessellations which may be obtained from the basic (3,3,3,3,3,3) tessellations by merging various sets of six equilateral triangles. (6, 6, 6) (3, 3, 3, 3, 6) No cutting is required in either case.

49. Section 2. Platonic and Archimedean Tilings 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.

50. Section 2. Platonic and Archimedean Tilings 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 1 x From we have x