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- Path: sparky!uunet!mcsun!uknet!acorn!armltd!dseal
- From: dseal@armltd.co.uk (David Seal)
- Newsgroups: rec.puzzles
- Subject: Re: Tiling a plane
- Message-ID: <4719@armltd.uucp>
- Date: 28 Jul 92 16:19:14 GMT
- References: <dfs.712246849@ro>
- Sender: dseal@armltd.uucp
- Distribution: rec
- Organization: A.R.M. Ltd, Swaffham Bulbeck, Cambs, UK
- Lines: 390
-
- In article <711892126.F00001@csource.oz.au> joe@csource.oz.au (Joe Slater) writes:
- >A plane can be tiled with three regular geometric shapes: squares, triangles and
- >hexagons. It can also be tiled with pairs of shapes: hexagons and triangles, or
- >octagons and squares.
- >
- >Are there any other pairs that will tile a plane? What about triplets? Answers
- >requiring fractal-like series to fill up gaps aren't counted.
-
- Yes, there are other pairs and some triplets. I'll look at tilings in which
- every vertex appears identical to every other one and all the tiles are
- regular polygons. For these, we can use the known formula for the angle of a
- regular polygon with N sides:
-
- angle = 180 * (1 - 2/N) degrees
-
- Suppose 3 polygons with A, B and C sides meet at each vertex. Then we know
- that the angles at the vertex must add up to 360 degrees:
-
- 180*(1-2/A) + 180*(1-2/B) + 180*(1-2/C) = 360
-
- Dividing through by 180:
-
- 3 - 2/A - 2/B - 2/C = 2
-
- Or:
-
- 2/A + 2/B + 2/C = 1
-
- Or:
-
- 1/A + 1/B + 1/C = 1/2
-
- So there is a possibility of such a tiling existing for any three integers
- whose reciprocals add up to 1/2. It is possible to enumerate such triplets
- quite easily. First, by reordering the polygons, we can assume that A <= B
- <= C. Now we know that 1/2 = 1/A + 1/B + 1/C <= 1/A + 1/A + 1/A = 3/A, which
- tells us that A <= 6. Since we know that A > 2, we can split into four
- cases:
-
- (1) A=3, 1/B + 1/C = 1/6. Working similarly, we can find that there are five
- possibilities:
-
- A B C
- -----------
- 3 7 42
- 3 8 24
- 3 9 18
- 3 10 15
- 3 12 12
-
- (2) A=4, 1/B + 1/C = 1/4. This gives us three possibilities:
-
- A B C
- -----------
- 4 5 20
- 4 6 12
- 4 8 8
-
- (3) A=5, 1/B + 1/C = 3/10. This gives one possibility:
-
- A B C
- -----------
- 5 5 10
-
- (4) A=6, 1/B + 1/C = 1/3. This also gives one possibility:
-
- A B C
- -----------
- 6 6 6
-
- There is then one further restriction, which you will find if you try to
- draw some of these patterns. Consider e.g. the A=3, B=8, C=24 triplet, and
- start trying to draw it. So we draw a triangle, an octagon and a 24-gon
- meeting at one vertex. The triangle and octagon share an edge: the vertex at
- the other end of that edge needs a 24-gon to make it complete. But now the
- remaining vertex of the triangle has two 24-gons adjacent to it: the tiling
- doesn't work. A diagram:
-
-
- /\ The top vertex doesn't work.
- 24-gon / \ 24-gon
- / \
- / \
- /Triangle\
- _------------_
- _- Octagon -_
-
- The same thing happens in general whenever one of A, B and C is odd: if the
- other two are not identical, we'll get a contradiction when we try to draw
- the tiling around the odd-sided polygon. So we get a rule:
-
- If any one of A, B and C is odd, the other two must be identical.
-
- Applying this rule to the 10 possibilities we discovered above, we find that
- only 4 remain possible:
-
- A B C
- -----------
- 3 12 12
- 4 6 12
- 4 8 8
- 6 6 6
-
- It's not immediately obvious that all four of these exist, but they do. The
- last of these is the regular tiling with hexagons; the second to last is the
- tiling with octagons and squares that you observed. The other two can be
- constructed from the hexagonal tiling:
-
- The (3,12,12) tiling:
- Start with the hexagonal tiling. Select a vertex and mark off points the
- same distance out along each of the three edges that come together at that
- vertex (make the distance less than half the edge length). If you join
- these three points together, you will get a small equilateral triangle
- centred on the original vertex, and each of the three adjacent hexagons
- has had a corner chopped off, becoming an irregular heptagon.
-
- Repeat this at each vertex, using the same distance every time. Each
- hexagon has all 6 corners chopped off to form 6 additional edges. Because
- the distance was less than half of the original edge length, some of the
- original edges remain, so each hexagon has become a dodecagon.
- Furthermore, the angles of these dodecagons are obviously all identical,
- and their edges are alternately the remains of an original hexagon edge
- and a triangle edge.
-
- If you vary the distance you mark off along each hexagon edge from 0 to
- half the original edge length, the triangle edges vary in length from 0 up
- to a positive amount and the remainders of the hexagon edges vary in
- length from a positive amount down to 0. For some particular choice of
- marked off distance, they will be the same, and you get a tiling with
- regular triangles and dodecagons. An attempt at a diagram:
-
- | | | |
- / \ / \ / \ / \
- _ _-----_ _-----_ _-----_ _-----_ _
- ----- ----- ----- ----- -----
- \ / \ / \ / \ / \ /
- | | | | |
- / \ / \ / \ / \ / \
- _-----_ _-----_ _-----_ _-----_ _-----_
- ----- ----- ----- -----
- \ / \ / \ / \ /
- | | | |
- / \ / \ / \ / \
- _ _-----_ _-----_ _-----_ _-----_ _
- ----- ----- ----- ----- -----
- \ / \ / \ / \ / \ /
- | | | | |
- / \ / \ / \ / \ / \
- _-----_ _-----_ _-----_ _-----_ _-----_
- ----- ----- ----- -----
- \ / \ / \ / \ /
- | | | |
-
- The (4,6,12) tiling:
- Again, start with the hexagonal tiling. This time, place a small square
- centrally on each of the edges, with two of its edges parallel to the
- original edge. Join each corner of each square to its nearest neighbour on
- another square: this will produce an irregular hexagon around each
- original vertex and an irregular dodecagon inside each original hexagon.
- However, the only way in which the hexagon and the dodecagon are irregular
- are that not all their edges are the same length, and if you choose the
- size of the squares correctly, this goes away and you get a tiling with
- regular squares, hexagons and dodecagons. Again, an attempt at a diagram:
-
- |__| |__| |__|
- \ / \ / \ / \
- \ / \ / \ / \
- / - __ - \ / - __ - \ / - __ - \ / -
- / / \ \ __ / / \ \ __ / / \ \ __ / /
- - / \ - - / \ - - / \ - - /
- \ / \ / \ / \
- \ __ / \ __ / \ __ / \
- | | | | | | |
- |__| |__| |__| |
- / \ / \ / \ /
- / \ / \ / \ /
- - \ / - __ - \ / - __ - \ / - __ - \
- \ \ __ / / \ \ __ / / \ \ __ / / \ \
- \ - - / \ - - / \ - - / \ -
- / \ / \ / \ /
- / \ __ / \ __ / \ __ /
- | | | | | | |
- | |__| |__| |__|
- \ / \ / \ / \
- \ / \ / \ / \
- / - __ - \ / - __ - \ / - __ - \ / -
- / / \ \ __ / / \ \ __ / / \ \ __ / /
- - / \ - - / \ - - / \ - - /
- \ / \ / \ / \
- \ __ / \ __ / \ __ / \
- | | | | | | |
-
- We can do exactly the same thing with four polygons (with A, B, C and D
- sides) meeting at each vertex. The equation corresponding to 1/A + 1/B + 1/C
- = 1/2 is now 1/A + 1/B + 1/C + 1/D = 1. Forcing A <= B <= C <= D, we get the
- following solutions:
-
- A B C D
- ---------------
- 3 3 4 12
- 3 3 6 6
- 3 4 4 6
- 4 4 4 4
-
- The criteria that ensure the tiling is possible are more complicated for
- these, due to the fact that there is more than one possible cyclic order of
- the polygons around the vertex. E.g. in the first case above, the two
- triangles may be adjacent to each other, separated on one side by the square
- and the dodecagon, or opposite each other, separated on one side by the
- square and on the other by the dodecagon. Neither of these leads to a
- possible tiling on its own, by similar arguments to those that established
- the "if one odd, the other two are identical" rule. A more complicated
- argument (which I will leave as an exercise) establishes that it also isn't
- possible to tile the plane using both types of vertex in the tiling.
- However, it *is* possible to tile the plane with triangles, squares and
- dodecagons if we allow other types of vertex as well. For instance, if we
- allow six triangles to meet at a vertex, we can derive such a tiling from
- the (4,6,12) tiling above by splitting each hexagon into 6 equilateral
- triangles.
-
- Similarly, the (3,3,6,6) vertex can have the two triangles adjacent to each
- other or opposite each other. Having them opposite each other leads to a
- completely regular tiling:
-
- _\____/__\____/__\____/_
- \ / \ / \ /
- \/ \/ \/
- /\ /\ /\
- __/__\____/__\____/__\__
- / \ / \ / \
- / \/ \/ \
- \ /\ /\ /
- _\____/__\____/__\____/_
- \ / \ / \ /
- \/ \/ \/
- /\ /\ /\
- __/__\____/__\____/__\__
- / \ / \ / \
-
- Less regular tilings with a mixture of the two types of vertex are also
- possible - e.g.:
-
- __/__\____/__\____/__\__
- \ / \ / \ /
- \/ \/ \/
- /\ /\ /\
- __/__\____/__\____/__\__
- \ / \ / \ /
- \/ \/ \/
- /\ /\ /\
- __/__\____/__\____/__\__
- \ / \ / \ /
- \/ \/ \/
- /\ /\ /\
- __/__\____/__\____/__\__
- \ / \ / \ /
-
- For (3,4,4,6), the two squares may be adjacent or opposite each other. A
- completely regular tiling is possible with the squares opposite each other -
- it's yet another modification of the hexagonal tiling:
-
- | _- \ / -_ | | _- \ / -_ |
- _| - \ / - |______| - \ / - |_
- \ \ ______ / / \ \ ______ / /
- \ _- | | -_ / \ _- | | -_ /
- \ - | | - / \ - | | - /
- / -_ | | _- \ / -_ | | _- \
- / - |______| - \ / - |______| - \
- _ / / \ \ ______ / / \ \ _
- | -_ / \ _- | | -_ / \ _- |
- | - / \ - | | - / \ - |
- | _- \ / -_ | | _- \ / -_ |
- _| - \ / - |______| - \ / - |_
- \ \ ______ / / \ \ ______ / /
- \ _- | | -_ / \ _- | | -_ /
- \ - | | - / \ - | | - /
- / -_ | | _- \ / -_ | | _- \
- / - |______| - \ / - |______| - \
- _ / / \ \ ______ / / \ \ _
- | -_ / \ _- | | -_ / \ _- |
- | - / \ - | | - / \ - |
-
- Again, there are also tilings that mix the two types - e.g.:
-
- | | | | | |
- -----+_ _+------+------+------+_ _+-----
- / --_ _-- \ / \ / --_ _-- \
- \ / /\ \ / \ / /\ \
- \/_ / \ _\/ \/_ / \ _\
- / --_ / \ _-- \ / --_ / \ _-- \
- / +------+ \ / +------+ \
- / \ \ / / \
- / \ _+------+_ / \
- --_ / \ _-- \ / --_ / \ _--
- /\ /\ \ / /\ /\
- / \ / \ _\/_ / \ / \
- / \ / \ _-- --_ / \ / \
- +------+------+------+ +------+------+------+
- | | | | | | | |
- | | | | | | | |
- | | | | | | | |
- +------+------+------+_ _+------+------+------+
- \ / \ / --_ _-- \ / \ /
- \ / \ / /\ \ / \ /
- _\/ \/_ / \ _\/ \/_
- -- \ / --_ / \ _-- \ / --
- \ / +------+ \ /
- \ / / \ \ /
- \ _+------+_ / \ _+------+_ /
- \ _-- \ / --_ / \ _-- \ / --_ /
- /\ \ / /\ /\ \ / /\
- / \ _\/_ / \ / \ _\/_ / \
- \ _-- --_ / \ / \ _-- --_ /
- -----+ +------+------+------+ +-----
- | | | | | |
-
- The (4,4,4,4) is of course the normal tiling with squares.
-
- For 5 tiles meeting at a vertex, the formula becomes 1/A + 1/B + 1/C + 1/D +
- 1/E = 3/2. With the usual re-ordering, this has two solutions:
-
- A B C D E
- -------------------
- 3 3 3 3 6
- 3 3 3 4 4
-
- Both of these lead to completely regular tilings. The completely regular
- (3,3,3,3,6) tiling is interesting because it is "skew" - i.e. it is not the
- same as its mirror image:
-
- /____\/____\/____\/ \/____\ ____ /____\
- \ / \ /\ /\ /\ /\ /
- \ / \ / \ / \ / \ / \ /
- __\/ \/____\ ____ /____\/____\/____\/
- /\ /\ /\ /\ / \ /\
- / \ / \ / \ / \ / \ / \
- /____\ ____ /____\/____\/____\/ \/____\
- \ /\ /\ / \ /\ /\ /
- \ / \ / \ / \ / \ / \ /
- __\/____\/____\/ \/____\ ____ /____\/__
- / \ /\ /\ /\ /\ /
- / \ / \ / \ / \ / \ /
- / \/____\ ____ /____\/____\/____\/
- \ /\ /\ /\ / \ /\
- \ / \ / \ / \ / \ / \
- __\ ____ /____\/____\/____\/ \/____\ __
- /\ /\ / \ /\ /\ /\
- / \ / \ / \ / \ / \ / \
- /____\/____\/ \/____\ ____ /____\/____\
- \ /\ /\ /\ /\ /
- \ / \ / \ / \ / \ /
- \/____\ ____ /____\/____\/____\/
- /\ /\ /\ / \ /\
- / \ / \ / \ / \ / \
- ____ /____\/____\/____\/ \/____\ ____
- \ /\ / \ /\ /\ /\ /
-
- There are two completely regular (3,3,3,4,4) tilings - I'll leave people to
- investigate these for themselves.
-
- With 6 polygons meeting at a vertex, the only solution is obviously that all
- six polygons are triangles, and we get the standard triangular tiling, and
- it is clearly impossible to produce such tilings with 7 or more polygons
- meeting at a vertex.
-
- So to summarise, the possibilities for these tilings (all polygons regular,
- all vertices appear exactly the same) are:
-
- A) A triangle and two dodecagons meet at each vertex.
- B) A square, a hexagon and a dodecagon meet at each vertex.
- C) A square and two octagons meet at each vertex.
- D) Three hexagons meet at each vertex.
- E) Two triangles and two hexagons meet at each vertex.
- F) A triangle, two squares and a hexagon meet at each vertex.
- G) Four squares meet at each vertex.
- H) Four triangles and a hexagon meet at each vertex.
- I) Three triangles and two squares meet at each vertex (two variants)
- J) Six triangles meet at each vertex.
-
- If you don't require all vertices to appear exactly the same, many more
- combinations are possible. E.g. it is possible to tile the plane with a
- combination of triangles, squares, hexagons and dodecagons, by taking tiling
- B) above and splitting some of the hexagons into six triangles (or some of
- the dodecagons into six triangles, six squares and a hexagon, which is also
- possible).
-
- David Seal
- dseal@armltd.co.uk
-
- All opinions are mine only...
-