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Tile a plane with three sided figures

  1. Jun 15, 2004 #1


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    I know one can tile a plane with three sided figures, four sided figures and six sided figures if each figure is identical. Are there any other numbers that would work?
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  3. Jun 15, 2004 #2


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    Actually, you can tile a plane with any even number of sides > 2, and 3 of course. I don't know if there are others.

    I will show an eight sided figure that can be used to tile a plane, and the rest follows.

    Imagine for now a regular octogon. Now, we take three consecutive sides of the octogon and flip them over to form a convex octogon which can fit into itself. This shape can be used to tile a plane. It follows that the same can be done for any greater even number of sides. This is just one example of a shape that can do it. There are infinitely many, of course.
  4. Jun 15, 2004 #3


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    Also there's...
    |__ |__ |__ |__ Ignore the dots (only to make spacing work)

    If that's impossible to comprehend, here, let me describe it...

    Imagine a pentagon, with 3 neighboring sides at right angles, and the other 2 sides equal to each other. Now stick 2 such pentagons together (see figure for clue) and you have a 7 sided (double-house) figure. You can tile a floor with these.

    In fact, the original pentagon (single-house) would itself work, as would, by extension, any number of such pentagons stuck to each other (n-house). All these figures have an odd number of sides.

    Since there is no requirement for regularity in the problem, this, along with vertigo's demonstration, shows that tile-solutions exist for all integers.
    Last edited: Jun 15, 2004
  5. Jun 23, 2004 #4


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    I think this gets harder if you add the requirement that the shapes be convex.

    Regarding regular polygons:
    Consider that the angle of a regular n-polygon is [tex]180-\frac{360}{n}[/tex], and for a regular polygon, the angle needs to be a divisor of [tex]180[/tex] or [tex]360[/tex]. Since any divisor of [tex]180[/tex] is also a divisor of [tex]360[/tex], it's sufficient to deal with divisors of [tex]360[/tex].

    Now, we know that the polygon will have 3 or more sides, so [tex]180-\frac{360}{n} \geq 60[/tex].
    Simultaneously we have [tex]180-\frac{360}{n} < 180[/tex].
    Now, we can list all divisors of 360:
    The divisors are
    1 2 3 4 5 6 8 9 10 12 15 18 20 24 30 36 40 45 60 72 90 120 180 360

    so the only possible divisors are :60,72,90, and 120.
    We know that 60 (hexagon), 90 (square) and 120 (triangle) are represented, so the only one left to check is
    but [tex]\frac{360}{108}[/tex] is not an integer, so there is no suitable regular polygon.

    Consequently, the ony regular polygons that tile the plane are triangles, squares, and hexagons.
  6. Jun 24, 2004 #5


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    check these tiles out : http://beloit.edu/~jungck/ ...Escher, I think.
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