# What's the best polygon for tiling the plane?

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?

• Are you allowed to use polygons of different shapes? – Jyrki Lahtonen Dec 11 '14 at 18:01
• @JyrkiLahtonen no – CLAUDE Dec 11 '14 at 18:05

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.

• Is there any proof? – CLAUDE Dec 11 '14 at 17:56
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. 