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Regular tessellations

  1. Aug 22, 2005 #1
    Show that the only regular figures that fills the plane are the triangle, the square and the hexagon.
     
  2. jcsd
  3. Aug 23, 2005 #2
    This can be done with algebra after the problem has been restated.

    Note that the angle of a corner of a regular n-gon equals (n-2).Pi/n.

    To fit the plane, k copies of a regular n-gon must be able to touch with their corners and therefore k angles should make up for a total arc of 2.Pi.
    So, k should be chosen such that

    k(n-2)Pi/n = 2Pi eq.
    k(n-2) = 2n (*) eq.
    k = 2n/(n-2).

    So, we must choose n such that n-2 | 2n.
    This looks already as if there are only few possibities. First note that n must be larger than 2. (Otherwise we don't even have a polygon).

    n=3 gives 1|6 which is true, and k = 6/1 = 6
    n=4 gives 2|8 which is true, and k = 8/2 = 4
    n=5 gives 3|10 which is NOT true
    n=6 gives 4|12 which is true, and k=12/4 = 3

    How does it go on?

    Well, the next quotient will be smaller than 3, so it must be 2, but then this would mean that just two corners n-gon fill an arc of 2Pi and this corner should be Pi, but this does not happen for a finite n.
    Therefore, there are no other regular n-gons that tesselate the plane.

    ---

    You can also look at solving the same equation (*) for n, we get:
    n = 2k/(k-2)

    So k should satisfy k-2 | 2k
    So either
    (i) k-2 = 1 or
    (ii) k-2 = 2 or
    (iii) k-2 is odd and k-2|k
    (iv) k-2 is even and (k-2)/2 | k

    Ad (i) k=3 and n=6
    Ad (ii) k=4 and n=4
    Ad (iii) k=2i+1 and 2i-1 | 2i+1. It is clear that if i>1 then (2i+1)/(2i-1) < 2, so this leaves no solutions
    Ad (iv) k=2i and (i-1)|2i eq. i=2 OR i=3 only. i=2 gives k=4 and we had this already. i=3 gives k=6 and n=3.

    This approach gives the same solutions.

    QED
     
  4. Aug 23, 2005 #3
    Thanks alot.
     
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