The question you are asking is by no means trivial. Fedorov found that there are exactly five 3-dimensional parallelohedra. You may be interested in the question.Įdit 2 : Let us consider $3D$ version. For example, (3,3,3,3,6) means there exist four equilateral triangles and one hexagon at every vertex.Įdit 1 : This is a question which I asked at mathoverflow. Also, you'll find some figures in the same page as above. You can find helpful comments in other's answer. Second, let's see the case we can use more than two distinct polygons and its copies to tessellate the plane. For example, you can divide a hexagon of (4) into two congruent pentagons. (6) Some pentagons with a special condition can tessellate the plane. (5) The fact (4) means that any quadrilateral can tessellate the plane because you can make a hexagon with three pairs of parallel edges using two copies of a quadrilatelral which is not a parallelogram. (4) A hexagon with three pairs of parallel edges can tessellate the plane. (3) The fact (2) means that any triangle can tessellate the plane because you can make a parallelogram using two copies of a triangle. (I found some figures here though the language is Japanese.) (2) You'll see that any parallelogram can tessellate the plane. (1) We can easily prove that there are only three regular polygons $(n=3,4,6)$ which tessellate the plane with one polygon. First, let's see the case that we use only one polygon and its copies to tessellate the plane.

Irregular tessellation definition