# Tessellations by Polygons

Oberkapfenberg Castle, Syria.

Relevant examples from Escher's work:

Recall that a polygon is a closed plane figure made by joining line segments. You might want to review the relevant material in Fundamental Concepts concerning polygons before reading this section.

## Some Basic Tessellations

The fundamental question we will discuss in this section is:

Which polygons tessellate?

More precisely, which polygons can be used as the only tile in a monohedral tessellation of the plane?

Before moving on, you may want to do the Tessellation Exploration: The Basics

The most common and simplest tessellation uses a square. Squares easily form horizontal strips:

Stacks of these strips cover a rectangular region and the pattern can clearly be extended to cover the entire plane. The same technique works with parallelograms, and so:

All parallelograms tessellate.

Special parallelograms such as rectangles, and rhombuses also tessellate.

Looking for other tessellating polygons is a complex problem, so we will organize the question by the number of sides in the polygon. The simplest polygons have three sides, so we begin with triangles:

All triangles tessellate.

To see this, take an arbitrary triangle and rotate it about the midpoint of one of its sides. The resulting parallelogram tessellates:

The picture works because all three corners (A, B, and C) of the triangle come together to make a 180° angle - a straight line. This property of triangles will be the foundation of our study of polygon tessellations, so we state it here:

The sum of angles of any triangle is 180°.

Moving up from triangles, we turn to four sided polygons, the quadrilaterals. Before continuing, try the Quadrilateral Tessellation Exploration.

Recall that a quadrilateral is a polygon with four sides.

The sum of angles in any quadrilateral is 360°

To prove, divide a quadrilateral into two triangles as shown:

Since the angle sum of any triangle is 180°, and there are two triangles, the angle sum of the quadrilateral is 180° + 180° = 360°. Taking a little more care with the argument, we have:

$\alpha_1 + \delta_1 + \gamma = 180^\circ$ and $\alpha_2 + \delta_2 + \beta = 180^\circ$.

Then

$360^\circ = \alpha_1 + \delta_1 + \gamma + \alpha_2 + \delta_2 + \beta  = \alpha_1 + \alpha_2 + \beta + \gamma + \delta_1 + \delta_2 = \angle A + \angle B + \angle C + \angle D$.

This division into triangles does not calculate the angle sum of the quadrilateral.

The point of all the letters is that the angles of the triangles make the angles of the quadrilateral, which would not work if the quadrilateral was divided as shown on the right.

We now turn to the main result of this section:

Begin with an arbitrary quadrilateral ABCD. Rotate by 180° about the midpoint of one of its sides, and then repeat using the midpoints of other sides to build up a tessellation.

The angles around each vertex are exactly the four angles of the original quadrilateral. Since the angle sum of the quadrilateral is 360°, the angles close up, the pattern has no gaps or overlaps, and the quadrilateral tessellates.

Recall from Fundamental Concepts that a convex shape has no dents. All triangles are convex, but there are non-convex quadrilaterals. The technique for tessellating with quadrilaterals works just as well for non-convex quadrilaterals:

It is worth noting that the general quadrilateral tessellation results in a wallpaper pattern with p2 symmetry group.

## Tessellations by Convex Polygons

Every shape of triangle can be used to tessellate the plane. Every shape of quadrilateral can be used to tessellate the plane. In both cases, the angle sum of the shape plays a key role. Since triangles have angle sum 180° and quadrilaterals have angle sum 360°, copies of one tile can fill out the 360° surrounding a vertex of the tessellation.

The next simplest shape after the three and four sided polygon is the five sided polygon: the pentagon. The angle sum of any pentagon is 540°, because we can divide the pentagon into three triangles:

Since each triangle has angle sum 180° the angle sum of the pentagon is 180° + 180° + 180° = 540°.

Rather than repeat the angle sum calculation for every possible number of sides, we look for a pattern. The angle sum of a triangle (3-gon) is 180°, the angle sum of a quadrilateral (4-gon) is 2x180°, and the angle sum of a pentagon is 3x180°. A general polygon with $n$ sides can be cut into $n-2$ triangles and so we have:

The sum of the angles of an $n$-gon is $(n-2)\times 180^\circ$.

Unlike the triangle and quadrilateral case, the pentagon's angle sum of 540° is not helpful when trying to fit a bunch of pentagons around a vertex. In fact, there are pentagons which do not tessellate the plane.

For example, the regular pentagon has five equal angles summing to 540°, so each angle of the regular pentagon is $\frac{540^\circ}{5} = 108^\circ$. Attempting to fit regular polygons together leads to one of the two pictures below:

Both situations have wedge shaped gaps that are too narrow to fit another regular pentagon. Thus, not every pentagon tessellates. On the other hand, some pentagons do tessellate, for example this house shaped pentagon:

The house pentagon has two right angles. Because those two angles sum to 180° they can fit along a line, and the other three angles sum to 360° (= 540° - 180°) and fit around a vertex.

Thus, some pentagons tessellate and some do not. The situation is the same for hexagons, but for polygons with more than six sides there is the following:

No convex polygon with seven or more sides can tessellate.

This remarkable fact is difficult to prove, but just within the scope of this book. However, the proof must wait until we develop a counting formula called the Euler characteristic, which will arise in our chapter on Non-Euclidean Geometry.

Nobody has seriously attempted to classify non-convex polygons which tessellate, because the list is quite likely to be too long and messy to describe by hand. However, there has been quite a lot of work towards classifying convex polygons which tessellate. Because we understand triangles and quadrilaterals, and know that above six sides there is no hope, the classification of convex polygons which tessellate comes down to two questions:

• Which convex pentagons tessellate?
• Which convex hexagons tessellate?

Question 2 was completely answered in 1918 by K. Reinhardt.[1] Reinhardt showed that there are only three types of convex hexagons which tessellate:

Reinhardt also addressed Question 1 and gave five types of pentagon which tessellate. In 1968, R. Kershner[2] found three new types, and claimed a proof that the eight known types were the complete list. A 1975 article by Martin Gardner[3] in Scientific American popularized the topic, and led to a surprising turn of events.

In fact Kershner's "proof" was incorrect. After reading the Scientific American article, a computer scientist, Richard James III, found a ninth type of convex pentagon that tessellates. Not long after that, Marjorie Rice, a San Diego homemaker with only a high school mathematics background, discovered four more types, and then a German mathematics student, Rolf Stein, discovered a fourteenth type in 1985.

Since 1985, no new types have been discovered, and many mathematicians believe that the list is finally complete. However, there is no well accepted proof of the classification, so it remains possible that there is a fifteenth or even many more types of convex pentagons that tessellate. Today, question 1 is an open problem, a problem whose solution is unknown.

Summary of Polygon Tessellations
Sides Angle Sum Tessellates?
3 180° Yes. All triangles tessellate.
4 360° Yes. All quadrilaterals tessellate.
5 540° Sometimes. There is a list of 14 types of convex pentagons which tessellate, but nobody knows if the list is complete.
6 720° Sometimes. There is a list of 3 types of convex hexagons which tessellate, and these are the only three.
7 & up 900°, 1080°, ... No convex $n$-gon tessellates for $n \geq 7$

## Tessellations by Regular Polygons

Recall that a regular polygon is a polygon whose sides are all the same length and whose angles all have the same measure. A regular $n$-gon has $n$ equal angles that sum to $(n-2)180^\circ$, so:

The corner angle of a regular $n$-gon is $\frac{(n-2)180^\circ}{n}$.

The table shows the corner angles for the first few regular polygons:

 Number of Sides Corner angle 3 4 5 6 7 8 9 10 11 12 60° 90° 108° 120° ~128.6° 135° 140° 144° ~147.3° 150°

### Regular Tessellations

Regular tessellation
A tessellation using one regular polygon tile, arranged so that edges match up.

Corners of the tiles need to fit together around a point, which means the corner angle of the regular polygon must evenly divide 360°. Since

$6 \times 60^\circ = 360^\circ$,

there is a regular tessellation using six triangles around each vertex. Since

$4 \times 90^\circ = 360^\circ$,

there is a regular tessellation using four squares around each vertex. And since

$3 \times 120^\circ = 360^\circ$

there is a regular tessellation using three hexagons around each vertex. We have already seen that the regular pentagon does not tessellate. A regular polygon with more than six sides has a corner angle larger than 120° (which is 360°/3) and smaller than 180° (which is 360°/2) so it cannot evenly divide 360°. We conclude:

There are three regular tessellations of the plane: by triangles, by squares, by hexagons.

A major goal of this book is to classify all possible regular tessellations. Apparently, the list of three regular tessellations of the plane is the complete answer. However, these three regular tessellations fit nicely into a much richer picture that only appears later when we study Non-Euclidean Geometry.

Tessellations using different kinds of regular polygon tiles are fascinating, and lend themselves to puzzles, games, and certainly tile flooring. Try the Pattern Block Exploration.

### Archimedean tessellations

This section is unfinished.

An archimedean tessellation using regular triangles and squares

## Notes

1. K. Reinhardt, Über die Zerlegung der Ebene in Polygone. (Inaugural-Disstertation, Univ. Frankfurt a.M.) R. Noske, Borna and Leipzig, 1918.
2. R. Kershner, On Paving the Plane, The American Mathematical Monthly 75, October 1968, pg. 839-844
3. Martin Gardner, Time Travel and Other Mathematical Bewilderments Ch. 13. W.H. Freeman, 1988