Math 230 Notes
Colouring the Vertices of an Equilateral Triangle
c
2003,
A. M. Dawes
Problem:
1. Each vertex (corner) of an equilateral triangle is coloured red or blue. We say that two
ways of colouring are equivalent under symmetries of the triangle if one can be obtained
from the other by rotating and/or flipping the triangle. Such equivalent colourings will not
be considered different in this problem. How many different ways are there to colour the
vertices? [Hint: how many equivalence classes?]
2. Same question, but now three colours (red, blue and green) are available. Consider two
cases:
(a) Two colourings are equivalent if one can be obtained from the other by rotation only;
(b) Two colourings are equivalent if one can be obtained from the other by rotation or
flipping (reflection).
3. Same question as (2), but n colours C1 , C2 , . . . , Cn are available.
Solution:
We give the general solution, after some observations:
If only one particular colour Ci is used, so that all vertices are coloured the same, then
rotation and/or reflection are irrelevant, and there is just one member of the equivalence
class of that colouring, represented by the diagram
C
•222 i
2
22
22
22
2
Ci C
• i
•
There are n such diagrams, one for each colour, representing n equivalence classes each with
one element. There are n ways to colour the triangle if the same colour must be used at each
vertex. (Note that there are n constant functions from the set {1, 2, 3} to {C1 , C2 , . . . , Cn }.)
When two different colours Ci and Cj are selected and both are used on the vertices, there
must be one vertex of one colour and two vertices of the other colour. The single-colour
vertex could be at the top, left, or right position of the triangle. However, all three ways to
colour the triangle can be obtained from each other by rotation (no flipping necessary); the
three diagrams below all belong to the same equivalence class.
1
C
•222 i
2
22
22
22
2
Cj C
j
•222
2
22
22
22
2
Ci <
2π/3
Cj
•
•
C
j
•222
2
22
22
22
2
Cj <
2π/3
Cj
•
•
•
Ci
•
The small curving arrows between the triangles show the rotation required to obtain the next
triangle (reading left to right). To specify an equivalence class, we need only state which
colour is used once (an n-way choice) and which different colour is used twice (an (n−1)-way
choice). There are n(n − 1) equivalence classes when exactly 2 of the n colours are used.
When three colours Ci , Cj and Ck are used, each vertex must have a different colour. In this
case it does make a difference whether we are permitted to flip the triangle over as well as
rotate it. We start with the situation where only rotation is permitted:
C
•222 i
2
22
22
22
2 Ck
Cj •
•
2π/3
<
C
•222 k
2
22
22
22
2 Cj
Ci •
•
C
2π/3
<
j
•222
2
22
22
22
2 Ci
Ck •
•
are all equivalent; each equivalence class has three members. The n(n − 1)(n − 2) ways to
choose colours for the top, left and right vertices are grouped into these equivalence classes,
so there are n(n−1)(n−2)/3 classes. That is, there are n(n−1)(n−2)/3 different colourings,
up to rotation.
In the case where flips are allowed as well, each of the 3 triangles shown above can be flipped,
say about a vertical axis, to obtain
C
22 j
•
2
2Ci
•22
Ck •
22
22
22
22
Cj
•
2π/3
<
Ci •
22
22
22
22
C
• k
2π/3
<
C
•222 k
2
22
22
22
2
Cj •
C
• i
So these colourings are all equivalent to the three above, and we find that the equivalence
classes have six members. Thus there are n(n − 1)(n − 2)/6 classes.
Conclusion:
If only rotations are allowed, there are n + n(n − 1) + n(n − 1)(n − 2)/3 non-equivalent
colourings.
If both rotations and flips are allowed, there are n+n(n−1)+n(n−1)(n−2)/6 non-equivalent
colourings.
Note: in addition to the general solution, you should be sure that you understand how to
get the special cases n = 2 and n = 3 separately, and check that your answers are consistent
with the general solution.
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