Sierpinski Triangle
What is the total area of the triangles remaining in the nth stage
of constructing a Sierpinski Triangle? Work out the dimension of
this fractal.
Age
16 to 18
Challenge level
Problem
The diagram shows the first three shapes in a sequence that goes on for ever and, in the limit, gives the Sierpinski triangle.
How many red triangles are there at Stage $n$? If the area of the red triangle at Stage $0$ is $1$, what is the total area of the red triangles at Stage $n$? How many white triangles are there at Stage $n$, that is how many triangles altogether have been removed? What is the total area of the triangles removed? What happens to these areas as $n \to \infty$? What is the dimension of this fractal?
What follows is information to help you answer these questions.
Image
The sequence starts with a red triangle. Each triangle in the sequence is formed from the previous one by removing, from the centres of all the red triangles, the equilateral triangles formed by joining the midpoints of the edges of the red triangles. If this process is continued indefinitely it produces a fractal called the Sierpinski triangle.
Now you are going to work out the dimension of this fractal. If you imagine moving about within this fractal then you have more choice of direction in which to go than if you were on a line and less choice of direction than in a square so you would expect the dimension of the fractal to be between $1$ and $2$.
We can break up the Sierpinski triangle into 3 self similar pieces $(n=3)$ then each can be magnified by a factor $m=2$ to give the entire triangle.
The formula for dimension $d$ is $n = m^d$ where $n$ is the number of self similar pieces and $m$ is the magnification factor.
NOTES AND BACKGROUND
If you break a line of length $1$ into self similar bits of length ${1\over m}$ there are $m^1$ bits and we say the dimension of the line is $1$.
If you break up a square of side $1$ into self similar squares with edge ${1\over m}$ then there are $m^2$ smaller squares and we say the dimension is $2$.
If you break up a cube of side $1$ into self similar cubes with edge ${1\over m}$ then there are $m^3$ smaller cubes and we say the dimension is 3.
In each case we say the magnification factor is $m$ meaning that we have to scale the lengths by a factor of $m$ to produce the original shape. So the formula is: $$\text{number of bits} = \text{(magnification factor})^d$$ where d is the dimension, i.e. $n = m^d$.
Getting Started
The fractal consists of the red triangles that remain if the process of removing the central 'quarter' of each red triangle is repeated indefinitely. Consider the first few stages, how many red triangles are there? How many triangles have been removed?
Student Solutions
Thank you Jeremy from Drexel University, Philadelphia, USA and Andrei, from Tudor Vianu National College, Bucharest, Romania for two more excellent solutions.
To solve this problem, first I made a table, and I filled it with the properties of the figures in the problem.
| Stage |
Number of
red triangles
|
Area
of red
triangles
|
Number of
white triangles
|
Area
of white
triangles
|
| $0$ | $1$ | $1$ | $0$ | $1-1$ |
| $1$ | $3$ | $3\over 4$ | $1$ | $1-{3\over 4}$ |
| $2$ | $3 \times3$ | ${3\over 4}\times {3\over 4}$ | $1+3$ | $1-\left({3\over 4}\right)^2$ |
From the analysis of the passage from one stage to the next, I made some observations:
- The number of red triangles at each stage is multiplied by three to give the number of red triangles at the next stage.
- The total area of the red triangles at each stage is multiplied by 3/4 to give the total area of the red triangles at the next stage.
- For each red triangle at stage $n$ one additional white triangle appears at stage $n+1$. Equivalently the number of white triangles added at each stage is three times the number of white triangles added at the previous stage.
- The total area of all the white triangles is [1 - (total area of red triangles)].
In the table below, I summarised the results obtained from
these observations, and I have also calculated the limits for $n
\to \infty$.[Jeremy gave a proof by induction for each of these
formulae]
| Stage |
Number
ofred triangles
|
Area
of red
triangles
|
Number
of white
triangles
|
Area
of white
triangles
|
| $n$ | $3^n$ | $\left(\frac{3}{4}\right)^n$ | $1 + 3 +3^2 ... +3^{n-1}= {3^n -1\over 2} $ | $1-\left(\frac{3}{4}\right)^n$ |
| $n \to \infty$ | $\infty$ | $0$ | $\infty$ | $1$ |
The total area of the white triangles (that is the triangles removed) is given by the series: $$\left({1\over 4}\right) + 3\left({1\over 4}\right)^2 + 3^2\left({1\over 4}\right)^3 +... 3^{n-1}\left({1\over4}\right)^n = 1 - \left({3\over 4}\right)^n$$ and this is as expected from the calculation of the total area of the red triangles.
In order to calculate the dimension $d$ from the formula $n=m^d$ , where $n$ is the number of self similar pieces and $m$ is the magnification factor, I see that for this problem $n=3$ and $m = 2$. So, $3 = 2^d$ which gives $\log 3 = \log\left(2^d\right)$ and hence $$d = {\log 3\over \log 2} =\log_2 3\approx 1.58 $$ I see that the dimension $d$ is between $1$ and $2$, as expected.
Teachers' Resources
If you break a line of length $1$ into self similar bits of length ${1\over m}$ there are $m^1$ bits and we say the dimension of the line is $1$.
If you break up a square of side $1$ into self similar squares with edge ${1\over m}$ then there are $m^2$ smaller squares and we say the dimension is $2$.
If you break up a cube of side$1$ into self similar cubes with edge ${1\over m}$ then there are $m^3$ smaller cubes and we say the dimension is $3$.
In each case we say the magnification factor is $m$ meaning that we have to scale the lengths by a factor of $m$ to produce the original shape. So the formula is: $$\text{number of bits} = \text{(magnification factor})^d$$ where $d$ is the dimension, i.e. $n = m^d$.