*Schläfli symbol*.
*infinitely* many regular tessellations of the hyperbolic plane. You can
determine whether {*n*,*k*} will be a tessellation of the Euclidean plane, the hyperbolic plane, or the elliptic plane by looking at the sum
1/*n* + 1/*k*. If the sum equals 1/2, as it does for the three tessellations mentioned above, then {*n*,*k*} is a Euclidean tessellation. If the sum is less than 1/2,
then the tessellation is hyperbolic; but if greater than 1/2, then elliptic.

*n*,*k*}, there are *k* regular polygons at each vertex. So the angle at each vertex is 360°/*k*. Since a regular *n*-gon has
*n* equal angles, each being 360°/*k*, therefore the angle sum is *n*360°/*k*.
*n*-gon is exactly (*n* - 2)180° in the Euclidean plane; less in hyperbolic; more in elliptic. Therefore, if *n*360°/*k* equals
(*n* - 2)180°, then {*n*,*k*} can only be Euclidean; if less, hyperbolic; and if more, elliptic. A little algebra (divide by *n*360° and add 1/*n*), and
you see that statement is becomes this: if 1/*n* + 1/*k* equals 1/2, then {*n*,*k*} can only be Euclidean; if less, hyperbolic; and if more, elliptic.

Variations of this {5,4}-tessellation are also available.

Other regular tessellations are available including {8,4}, {4,8}. {6,6}, {3,12}, and {12,3}.

*n*,*k*} gives rise to a
quasiregular tessellation quasi-{*n*,*k*} by connecting the
midpoints of the edges of the regular tessellation. In the Euclidean plane
there are just two quasiregular tessellations: quasi-{3,6}
arises from both {3,6} and {6,3}, while
quasi-{4,4} comes from {4,4}. (Of course,
quasi-{*n*,*n*} is the same as {*n*,4}.)

Here's a variation of it where the squares aren't colored, but pentagrams of various colors are placed in the pentagons (the pentagons don't show, either).

Other quasiregular tessellations.

August 1994; Dec 1998

David E. Joyce

Department of Mathematics and Computer Science

Clark University

Worcester, MA 01610

The address of this file is http://aleph0.clarku.edu/~djoyce/poincare/poincare.html