Tile a plane with three sided figures

[mee]

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?

 

[verty]

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.

 

[Gokul43201]

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.

 

[NateTG]

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.

 

[Gokul43201]

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