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The Towers of Riga

Discover the mathematical solution behind fitting the twelve ancient towers of Riga into a modern museum design. Can these towers actually fit together? Find out in this intriguing exploration of mathematics and architecture.

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The Towers of Riga

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  1. The Towers of Riga Chiu Chang Mathematics Education Foundation President Wen-Hsien SUN

  2. There are twelve ancient towers in Riga. The base of each consists of three regular hexagons each sharing one side with each of the other two. On these regular hexagons are right prisms of varying heights, as indicated by the numbers in Figure 1 below.

  3. 5 3 2 2 2 6 1 5 2 1 5 2 6 1 1 1 1 5 2 6 1 6 6 5 5 4 3 3 2 1 6 6 6 5 4 4 D Figure 1 C A B G F H E I J L K

  4. A legend says that the gods used magic to gather these twelve towers to the center of the town, and stored them in a large hexagonal box with a flat top and bottom and no spaces in between.

  5. 6 5 2 Naming the Pieces • The pieces are named by, starting at the shortest section, naming the height of each section in counterclockwise order. For example, the piece on the left would be referred to as (2, 5, 6).

  6. 2 1 1 Naming the Pieces • If there are two identical sections that are shortest, they are not separated, for example the bottom right piece would be referred to as (1, 1, 2)

  7. 7 7 7 7 7 7 7 7 7 7 7 7 4 7 7 7 7 7 7 7 • It is decided that the twelve towers are to be combined into a modern museum of height 7, with the pattern of the base as shown in either diagram of Figure 2. The pink central hexagon represents the elevator shaft.

  8. Figure 2

  9. Clearly, six of the towers will have to be upside down, so that their flat bases form part of the flat top of the museum. Assuming that the practical difficulties can be overcome, the natural question is whether the towers actually fit together. In other words, is there a mathematical solution?

  10. 7 7 7 7 7 7 If both the top and the bottom of the apartment block follow the same one of the two patterns in Figure 2, then the twelve towers must form six complementary pairs, such as the (1,1,1) and the (6,6,6).

  11. 1 6 6 1 1 6 6 6 1 1 6 1 1 1 1 6 6 6 + 7 7 7 =

  12. However, we can see quickly that this is not the case. For instance, we have a (4,6,6) but no (1,1,3).

  13. 4 4 4 6 6 6 6 6 6 1 1 1 1 1 1 + 3 3 3 7 7 7 =

  14. 5 3 2 2 2 1 6 2 5 1 1 1 2 1 6 5 1 1 1 1 5 2 6 6 6 5 5 4 3 3 2 1 3 6 6 6 5 4 4 D C A B G F H E I J L K

  15. Thus we may assume that the base of the apartment block follows the pattern on the left of Figure 2, and the top follows that on the right.

  16. 7 7 7 7 7 7 7 7 7 7 7 7 4 7 7 7 7 7 7 7

  17. This implies that the twelve towers form six overlapping pairs, one right side up and the other upside down. Two of the hexagonal prisms of the latter are resting on two of those of the former, so that the two vertical neighbors have a combined height of 7.

  18. 7 7 7 7 7 7 7 7 7 7 7 7 4 7 7 7 7 7 7 7

  19. 4 4 4 4 6 6 6 6 6 6 60° 120° 60° 1 1 1 6 1 1 1 1 1 6 6 1 6 1 6 1 1 1 1 1 1 6 1 1 1 1 1 1 6 6 6 1 1 1 1 1 7 1 7 7 6 1 6 7 6 4 2. A+L 1. A+J A 1. 2. + + = =

  20. 60° 60° 60° 2 2 2 1 1 1 1 1 1 6 6 6 6 5 5 5 5 5 5 5 5 3 3 3 3 3 180° 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 7 2 1 1 7 7 180° 7 6 6 6 6 2 2 2 2 2 2 2 2 2 2 1. B+G 2. B+H B 1. 2. + + = =

  21. 4 4 4 4 4 6 6 6 6 6 6 120° 120° 60° 60° 2 2 1 1 1 1 6 6 6 6 2 5 5 5 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 7 2 2 2 2 1 7 7 180° 180° 6 7 6 2 2 2 2 2 4 4 4 4 4 1. B+G 2. B+H B 3. B+I 4. B+J 3. 4. + + = =

  22. 120° 2 6 6 1 6 6 1 6 6 2 1 1 1 1 1 1 1 1 1 1 2 2 2 2 7 180° 7 6 6 6 6 2 1. B+G 2. B+H B 3. B+I 5. B+L 4. B+J 5. + =

  23. 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 120° 60° 60° 60° 3 2 1 6 6 6 6 6 6 6 6 5 5 5 5 3 2 7 7 2 7 7 180° 180° 6 6 6 6 6 2 2 2 4 4 4 4 1. C+G 2. C+J C 1. 2. + + = =

  24. 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 3 3 3 3 3 3 3 3 3 3 3 60° 60° 60° 120° 3 2 1 6 6 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 2 7 7 180° 180° 7 7 4 4 4 4 4 4 4 4 4 1. C+G 2. C+J C 3. C+I 3.(a) 3.(b) + + = =

  25. 5 60° 120° 60° 120° 2 1 6 6 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 180° 180° 1 1 1 1 1 1 1 1 1 1 2 2 7 7 2 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 7 7 D 2. D+G 1. D+I 1. 2.(a) + + = =

  26. 5 60° 60° 2 1 6 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 180° 180° 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 4 7 7 6 2 2 2 2 2 2 2 2 2 2 2 2 7 7 D 2. D+G 1. D+I 2.(b) 2.(c) + + = =

  27. 120° 4 4 180° 3 3 1 1 1 1 1 1 7 4 4 4 4 4 6 6 3 3 3 3 7 6 6 6 6 6 4 4 4 E E+J + =

  28. 2 5 2 2 2 5 5 2 2 5 5 2 2 5 2 5 5 2 5 2 2 + 7 2 7 5 F F+K =

  29. 5 3 2 2 2 2 6 5 1 6 1 2 6 6 5 1 1 1 1 2 5 6 6 1 6 6 6 6 5 5 5 5 4 3 3 2 2 2 1 6 6 6 6 6 6 6 6 6 6 6 5 5 5 4 4 4 4 4 4 4 G J K G G J I I J I L L J H

  30. 5 3 2 2 2 6 1 6 5 2 1 6 1 2 6 5 1 1 1 6 2 5 6 1 6 6 6 6 5 5 5 5 4 3 3 2 2 2 1 6 6 6 6 6 5 5 5 4 4 4 4 G J K L G G I L I I H

  31. 5 3 2 2 2 6 1 5 2 1 6 2 5 1 1 1 1 6 5 1 2 6 6 6 5 5 5 4 3 3 2 2 1 6 6 6 6 5 5 4 4 4 J K L H G G I I

  32. A+L 1 6 6 1 1 6 1 6 6 1 1 6 1 6 1 1 6 6 + 7 1 7 6 (1, 1, 1)+(6, 6, 6)=(6, 1) =

  33. 120° 2 1 1 6 5 180° 3 7 2 2 2 5 5 5 1 1 1 1 6 6 6 7 1 1 1 3 3 3 3 B+H (1, 1, 2)+(3, 5, 6)=(3, 1) + =

  34. E+J 120° 4 180° 3 1 1 1 1 1 7 4 4 4 4 4 6 6 3 3 3 3 7 6 6 6 6 6 4 4 4 (1, 3, 4)+(4, 6, 6)=(6, 4) + =

  35. F+K 2 5 5 5 2 2 5 2 5 2 2 5 2 5 5 2 5 2 + 7 2 7 5 (2, 2, 2)+(5, 5, 5)=(5, 2) =

  36. 5 3 2 2 1 1 6 5 2 7 7 7 7 1 4 7 2 1 7 6 7 7 6 5 6 5 3 4 Now we get the following accessories (3, 1) (5, 2) (6, 4) (6, 1) D G I C

  37. We identify (2,2,2)+(5,5,5), (1,3,4)+(4,6,6), (1,1,2)+(3,5,6) and (1,1,1)+(6,6,6) with the dominoes (2,5), (4,6), (1,3) and (1,6), respectively.

  38. Each of the dominoes (4,6) and (1,3) links a (1,6) with a (1,6). 60° 120° 7 7 7 7 7 7 7 7 6 7 7 7 7 7 1 1 1 1 4 4 4 7 7 7 7 7 1 7 7 7 7 7 7 6 6 6 6 6 7 1 1 1 7 6 7 7 7 7 1 3 3 + + =

  39. 7 7 7 7 6 7 7 7 2 7 7 7 5 1 • In order to get the (2,5) into the domino ring, we need at least one domino in which one of the numbers is 2 or 5, and the other is 1 or 6. With this in mind, we now examine the remaining four towers. There are two possible couplings.

  40. 5 3 2 2 1 1 6 6 5 5 2 2 6 6 5 5 4 4 (i) C+I、D+G (ii) C+G、D+I G G I I

  41. (i) 5 3 2 2 1 1 6 6 5 5 2 2 2 5 5 1 1 6 6 3 7 7 7 7 7 4 4 7 7 7 7 7 D+G C+I + + False

  42. (ii) 5 3 2 2 1 1 6 6 5 5 2 2 5 3 4 7 7 4 7 7 D+I C+G + + False

  43. Each of the dominoes (4,6) and (1,6) links a (1,3) with a (3,4).[Method 1] 60° 60° 7 1 7 7 1 7 6 3 7 7 7 180° 180° 180° 1 7 7 7 7 7 7 7 7 7 7 7 3 3 3 7 7 4 4 4 7 7 7 7 1 7 7 6 6 6 4 7 7 7 6 6 6 7 1 7 7 6 7 7 6 4 7 7 1 3 3 + + =

  44. Each of the dominoes (4,6) and (1,6) links a (1,3) with a (3,4).[Method 2] 60° 60° 7 1 7 7 1 7 6 3 180° 180° 180° 7 7 7 1 7 7 7 7 7 7 7 7 7 7 7 7 4 4 4 7 7 7 7 1 1 7 7 7 4 6 6 6 7 7 7 7 7 6 6 6 1 7 3 3 3 6 7 7 6 7 4 7 3 3 + + =

  45. 7 7 7 7 7 7 7 7 7 7 2 7 7 7 7 7 4 4 7 5 7 7 3 3 • In order to get the (2,5) into the domino ring, we need at least one domino in which one of the numbers is 2 or 5, and the other is 3 or 4. With this in mind, we now examine the remaining four towers. There are two possible couplings.

  46. 5 3 2 2 1 1 6 6 5 5 2 2 6 6 5 5 4 4 (i) C+I、D+G (ii) C+G、D+I G G I I

  47. (i) 5 3 2 2 1 1 6 6 5 5 2 2 2 5 5 1 1 6 6 3 7 7 7 7 7 4 4 7 7 7 7 7 D+G C+I + + False

  48. (ii) 5 3 2 2 1 1 6 6 5 5 2 2 5 3 4 7 7 4 7 7 D+I C+G + + OK!

  49. Method 1 7 7 2 2 2 2 5 5 5 5 7 7 = 7 7 7 7 7 3 3 3 3 2 7 7 7 7 7 4 4 4 4 4 4 4 7 7 7 7 7 7 7 7 7 7 5 7 7 7 7 7 3 3 3 7 7 7 7 7 7 7 7 2 2 2 7 7 7 5 5 5 + + 7 7 7 7 7 7 7 7 7 7 + =

  50. Method 1 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 4 4 7 7 7 7 7 7 7 7 7 7 7 7 7 7

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