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## 'Sierpinski Triangle' printed from http://nrich.maths.org/

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.

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. $n
= m^d$.