**Tiling the Plane with Convex Pentagons**

Prepared by:

Joseph Malkevitch

Department of Mathematics and Computer Studies

York College (CUNY)

Jamaica, New York

email:

__malkevitch@york.cuny.edu__

web page:

__http://york.cuny.edu/~malk__

A much looked at question in elementary geometry has been what convex pentagons tile the plane. This problem was recently (2015) revitalized when a new convex pentagonal tile for the plane was discovered by a team lead by Casey Mann based on a machine search.

Discussion set off by the new discovery focused attention on the fact that the way convex pentagons tile the plane has been organized in a way that allows several families of tiles to have things in common. A summaries of what was known prior to the new discovery are available here:

http://www.mathpuzzle.com/tilepent.html

and here where the new discovery is also mentioned:

https://en.wikipedia.org/wiki/Pentagonal_tiling

Here is an account of the new discovery:

http://www.npr.org/sections/thetwo-way/2015/08/14/432015615/with-discovery-3-scientists-chip-away-at-an-unsolvable-math-problem

Perhaps the following ideas may be of use in organizing ideas about what convex pentagons tile the plane as well as trying to make progress on obtaining a complete list of such polygons.

It may be that a "point of view" that I have been trying to promote may be useful to those who are working on the enumeration of convex pentagonal tiles of the plane. Some years ago Branko Grünbaum wrote about an innovative and clever idea for classifying convex 4-gons. My contribution to this, via some additional work by an EdD student of mine at Teachers College at Columbia, was to think of what Branko had done in a partitions framework, though the "heavy lifting" was Branko's original idea.

For pentagons, the idea would be: classify convex pentagons by a pair of partitions, one for sides and one for angles. There are 7 partitions of 5, so a priori there would 49 possible types of pentagons not all of which might exist. For example, {3, 2};(3, 1, 1) would denote a convex pentagon with 3 equal sides of one length and two of another length and also, 3 equal angles of one measure, and two other angles of different measure, and these two angles would not be of the same measure. So the idea would be for each of the potential types (check if it exists) and if it does try to show that there are convex pentagonal tiles of this type or show there are none.

For convex 4-gons Branko showed for off diagonal types there was an unexpected "reciprocity" - if type i;j did or did not exist the same was try for type j;i. (Branko made a small mistake in the total list of 4-gons but this mistake (corrected by my student) was on the diagonal so did not affect his theorem but only the total count.) My student (Orlando Alonso) also extended the partitions approach to additional labels for angles regarding whether angles were acute, right or obtuse, and carried out the enumeration for 4-gons.

To the best my knowledge other than a conjecture about extending the idea of "reciprocity" to convex polygons with more sides (in the partitions case) made by me and Alonso (Branko's original form of the conjecture had a counterexample, but not when restricted to a partitions framework), I am not sure anything has been done for the case of pentagons.

A joint paper by Orlando and me for teachers about these ideas is:

Alonso, Orlando B., and Joseph Malkevitch. "Classifying Triangles and Quadrilaterals." Mathematics Teacher 106, no. 7 (2013): 541-548.

and another "expository paper"

Alonso, Orlando B., and Joseph Malkevitch. "Enumeration via Partitions." Consortium 98 (2010): 17-21.

Reference to Branko's original paper is:

Grünbaum, Branko. "The angle-side reciprocity of quadrangles." Geombinatorics 4 (1995): 115-118.

and Orlando has some papers about non-convex quadrilaterals in Geombinatorics.

Orlando also, at my suggestion, for convex quadrilaterals, looked at "compositions." These are ordered partitions, so {3,1,1} would correspond to the compositions {3, 1, 1}, {1, 3, 1}, and {1, 1, 3} where one can now think of what is being distinguished as the sizes of the sides. {3, 1, 1} would mean the biggest side lengths were all equal while {1, 1, 3} would mean the smallest sides lengths were equal and similarly for angles. Where as it is not easy to count partitions, compositions of n are easy to count. For the value n there are 2^(n-1) compositions. So for 5, there are 16. So one could, with somewhat more work, do what I described above using compositions rather than partitions.

For those actively looking at tilings by congruent convex pentagons, perhaps these ideas might "focus" attempts to find new types of tiling pentagons and/or show that there are no more.