Tile a plane with three sided figures

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Tiling a plane can be achieved with various polygonal shapes, specifically those with three or more sides. It is established that any even-sided polygon greater than two can tile a plane, along with triangles. An example provided is a modified octagon that can tile a plane by flipping three consecutive sides to create a convex shape. Additionally, irregular polygons, such as certain configurations of pentagons, can also tile a plane, demonstrating that solutions exist for all integers. However, when focusing on regular polygons, only triangles, squares, and hexagons can tile the plane due to the requirement that the internal angles must be divisors of 360 degrees. The analysis of divisors confirms that no regular polygon with 72 degrees can tile the plane, solidifying the conclusion that the only regular polygons capable of tiling are those previously mentioned.
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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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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.
 
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.
 
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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 180-\frac{360}{n}, and for a regular polygon, the angle needs to be a divisor of 180 or 360. Since any divisor of 180 is also a divisor of 360, it's sufficient to deal with divisors of 360.

Now, we know that the polygon will have 3 or more sides, so 180-\frac{360}{n} \geq 60.
Simultaneously we have 180-\frac{360}{n} < 180.
Now, we can list all divisors of 360:
360=2^3*3^2*5
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
72:
72=180-\frac{360}{n}
\frac{360}{108}=n
but \frac{360}{108} 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.
 
check these tiles out : http://beloit.edu/~jungck/ ...Escher, I think.
 
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