## Page 1 of 3

12 At Right Angles | Vol. 3, No. 1, March 2014

feature

I

n this note we show how the semi-regular tessellations can be

enumerated. We describe only the approach and give a partial

solution. First we recall the de�inition: A semi-regular tessellation is a

�illing of the plane with regular polygons of two or more kinds, such that

the polygons with a given number of sides are congruent copies of one

another, and the pattern of placement of the polygons is the same at every

vertex of the tessellation. Figure 1 shows two semi-regular tessellations.

Pattern I is made up of squares and regular octagons. Going around each

vertex we meet a square and two octagons, so we associate the tuple

(4,8,8) with the pattern. Pattern II is made up of equilateral triangles,

squares and regular hexagons, and we associate the tuple (3,4,6,4) with

the pattern.

Pattern I: (4,8,8) Pattern II: (3,4,6,4)

Figure 1.

Source: http://en.wikipedia.org/wiki/Tiling_by_regular_polygons

Enumeration of

Semi-regular

Tessellations

Keywords: Tessellation, semi-regular, tiling

## Page 2 of 3

At Right Angles | Vol. 3, No. 1, March 2014 13

We now describe here how all such tuples can be

enumerated, but we leave the task for you to

complete. Let the tuple be (nnnn�, nnnn�, ... , nnnn�) where

each nnnn� is a positive integer. Since each polygon

has at least three sides, we have nnnn� ≥ 3 for every iiii.

We start by showing that kkkk kk kk. That is, there can

be no more than six polygons meeting at each

vertex.

Recall that the internal angle of a regular nnnn-sided

polygon is 180∘ − 3kk0∘

/nnnn. Since the total angle at

each vertex is 3kk0∘

, it follows that

�180∘ − ���∘

�� � + �180∘ − ���∘

�� � + ⋯ + �180∘ − ���∘

�� � = 3kk0∘

.

(1)

Dividing through by 180∘ we get:

�1 − 2

nnnn�

� + �1 − 2

nnnn�

� + ⋯ + �1 − 2

nnnn�

� = 2. (2)

There are kkkk bracketed terms. Opening the

brackets and simplifying, we get:

1

nnnn�

+

1

nnnn�

+ ⋯ +

1

nnnn�

= kkkk

2 − 1. (3)

We need to �ind tuples (nnnn�, nnnn�, ... , nnnn�) of positive

integers satisfying (3) and the condition that

nnnn� ≥ 3 for all iiii. This condition implies that

1/nnnn� kk 1/3 for every iiii, and hence that:

1

nnnn�

+

1

nnnn�

+ ⋯ +

1

nnnn�

kk

kkkk

3. (4)

From (3) and (4) we deduce:

kkkk

2 − 1 kk

kkkk

3, ∴

kkkk

2 − kkkk

3 kk 1, ∴

kkkk

kk kk 1, (5)

which leads to kkkk kk kk, as claimed. On the other

hand, kkkk ≥ 3, for we cannot have less than three

polygons meeting at a vertex. So kkkk kk kk3, kk, kk, kkkk.

Thus, kkkk can take just four possible values, and we

can enumerate the solutions of (3) by proceeding

case by case. For now we examine only the case

kkkk = 3, and the leave the others for you.

For convenience rename nnnn�, nnnn�, nnnn� as aaaa, aaaa, aaaa. There

is no harm in relabeling them so that aaaa kk aaaa kk aaaa.

Since kkkk/2 − 1 = 3/2 − 1 = 1/2, the system we

have to solve is:

1

aaaa

+

1

aaaa +

1

aaaa = 1

2

, 3 kk aaaa kk aaaa kk aaaa. (6)

From aaaa kk aaaa kk aaaa we get

1

aaaa

+

1

aaaa +

1

aaaa

kk

1

aaaa

+

1

aaaa

+

1

aaaa = 3

aaaa. (7)

Therefore 3/aaaa ≥ 1/2, and aaaa kk kk. Hence

aaaa kk kkkk kkk kkk kkkk. We now take up each possibility in

turn.

aaaa = aaaa We have: 1/aaaa aaaaaaa aaaaaaaaaaaaa.

Hence aaaaaaaa aaaaaaaa a aaaaaaa. Adding 3kk to both

sides to achieve a factorization we get

(aaaa aaaaaaaaa aaaaaaaa. From the factorization

of 3kk we infer that (aaaa aaaa aaaa aaaa is one of the

pairs (1, 3kk), (2, 18), (kk, 9), (kk, kk). Hence

(aaaaa aaaaa is one of the following: (7, kk2), (8, 2kk),

(9, 18), (10, 1kk), (12, 12).

aaaa = aaaa In the same way we get:

1/aaaa aaaaaaa aaaaaaaaaaaaaa, hence

aaaaaaaa aaaaaaaa a aaaaaaa, which yields

(aaaa aaaaaaaaa aaaaaaaa. So (aaaa aaaa aaaa aaaa is one

of the pairs (1, 1kk), (2, 8), (kk, kk), implying

that (aaaaa aaaaa is one of the following: (kk, 20),

(kk, 12), (8, 8).

aaaa = aaaa This time we are led to the equation

3aaaaaaaa aaaaaaaa a aaaaaaa. Solving this the same

way (and remembering that aaaa kk aaaa), we �ind

that (aaaaa aaaaaaaaaaaaa. (Details omitted.)

aaaa = aaaa This time we are led to the equation

aaaaaaaa aaaaaaa a aaaaaaa. Solving this the same way

(and remembering that aaaa kk aaaa), we �ind that

(aaaaa aaaaaaaaaaaaa. (Details omitted.)

Hence (aaaa, aaaa, aaaa) is one of the following triples:

(3, 7, kk2), (3, 8, 2kk), (3, 9, 18), (3, 10, 1kk),

(3, 12, 12), (kk, kk, 20), (kk, kk, 12), (kk, 8, 8), (kk, kk, 10),

(kk, kk, kk). Of these, the last is a regular tessellation,

as there is just one type of polygon (a regular

hexagon).

We now examine the other triples in the list. We

shall eliminate many of them using a clever parity

argument. Here is the claim: If any one out of

aaaa, aaaa, aaaa is odd, then the other two numbers must be

equal. The proof may be grasped by examining

any triple with an odd entry, say (3, 10, 1kk). The

numbers tell us that around each vertex there is

an equilateral triangle, a regular decagon and a

regular 1kk-sided polygon. Focusing attention on

the triangle and going around its edges, we see

that the decagon and 1kk-sided polygon must come

in alternation. But this is impossible, since the

triangle has an odd number of sides!

2

## Page 3 of 3

This argument extends to all triples with one odd

number and two other numbers which are

unequal. After eliminating the triples which do

not conform to the rule, we �ind that only these

remain: (3, 12, 12), (4, 6, 12) and (4, 8, 8). Each of

these corresponds to a genuine semi-regular

tessellation.

The arguments for the cases kkkk kk 4, kk, 6 may be

conducted along similar lines, though there are

many more subtleties involved. But for now, we

leave these to the reader. For further reading

please refer to the following:

http://en.wikipedia.org/wiki/Tiling_by_regular_polygons

http://www.mathsisfun.com/geometry/tessellation.html

http://mathforum.org/sum95/suzanne/whattess.html

3 14 At Right Angles | Vol. 3, No. 1, March 2014

The COMMUNITY MATHEMATICS CENTRE (CoMaC) is an outreach sector of the Rishi Valley Education Centre

(AP). It holds workshops in the teaching of mathematics and undertakes preparation of teaching materials for

State Governments, schools and NGOs. CoMaC may be contacted at comm.math.centre@gmail.com.

It was a peculiar looking

board, with all sorts of

odd shapes carved out of

a wooden plank. I stood

there, examining it for a

while, trying to understand

its place in an exhibition

on the ‘Mathematics of

Planet Earth’. Observing

my confusion, one of the

organisers walked up to

me. “Do you see the little

sticks with the red bulb

at one end that those kids are holding? At the other end of it is a thin metallic strip. Do you think

you can rotate it 180 degrees within each of these shapes carved out of the board? Naturally, you are

not permitted to lift the rod o the board at any point in time, as our enterprising young friend is

attempting here.” he said, walking a couple of steps to explain to the game to the child in question.

While patiently waiting for my turn, I thought about the task at hand. It seemed unlikely. The rod

was rigid, unbending and the shapes, although they began fairly regularly with the circle, soon became

strange. When my turn fi nally came, I picked up the rod and gradually tried to manoeuvre it this way

and that. With some e ort, however, I fi nally managed to wiggle the rod around and rotate it the

desired way within each of the fi gures cut out. Feeling rather pleased with myself for having worked

out the brain teaser, I looked up, only to fi nd our organiser friend observing my handiwork. “Not

bad!”, he exclaimed. “Do you see this triangle with the sides having an ‘outward bulge’? Say that has

area A. Then this regular triangle alongside it should, intuitively, have area slightly less than A. What

about the area of this ‘inward bulging’ triangle adjacent to this? That will have even smaller area!” I

could see where he was going with this line of reasoning: Would this ever stop? Is there a ‘smallest’ set

(in terms of area) within which you can perform this rotation? I thought about it for a while before I

ventured a guess. Why don’t you try the same, and then turn to page 35 to check if our ideas match!

A needle - ing problem