What's the best polygon for tiling the plane?

Question

We want to cover the whole plane by tiles, shaped as a polygon with equal-length sides, such that there is not overlapping and any gap (Note that all the tiles are similar to each other). which polygon has the maximum number of sides that can do that?

Answer 1

If you want to use only one shape that is a regular n-gon, hexagon is as good as it gets.

EDIT: Proof from here http://mathandmultimedia.com/2011/06/04/regular-tessellations/

The angle sum of a polygon with p sides is 180(p-2). This means that each interior angle of a regular polygon measures $\frac{180(p-2)}{p}$. The number of polygons meeting at a point is q. The product is therefore

$\frac{180(p-2)}{p}(q) = 360$

which simplifies to pq - 2q = 2p. Using Simon’s Favorite Factoring Trick, we add 4 to both sides giving us pq - 2q + 4= 2p + 4. Factoring and simplifying, we have q(p - 2) - 2(p - 2) = 4, which is equivalent to (p - 2) (q - 2) = 4. Observe that the only possible values for {p,q} are {4,4} (squares), {6,3} (regular hexagons), or {3,6} (equilateral triangles). This means that these are the only regular tessellations possible which is what we want to prove.

Answer 2

I would like to add that if you do not require that the polygon be regular, there is no maximum number of sides to tile the plane. For example, if you take $7$ hexagons, one in the center with $6$ bordering it, the polygon enclosing them has $18$ edges of equal lengths and still tiles the plane. You can then take $7$ of those shapes to get a polygon with $54$ equal-length sides, that tiles the plane. You can repeat this process indefinitely.

The limit of this process (with resizing at each iteration) is a fractal called the Gosper Island, which also tiles the plane.