Teachers' Notes Hint Solution

Von Koch Curve

Thank you Shaun from Nottingham High School and Andrei from Tudor Vianu National College, Bucharest, Romania for your solutions. Both Shaun and Andrei used essentially the same method.

To solve this problem I made a table with the first $3$ stages and then I determined analytical expressions for the quantities in the table as a function on $n$:

Stage	Number of sides	Length of curve	Area contained inside the curve
$0$	$3$	$3$	${\sqrt 3\over 4}$
$1$	$3\times 4=12$	$3\times{4\over 3}$	${\sqrt 3\over 4}\left[1+3\left({1\over 9}\right)\right]$
$2$	$3\times 4^2=48$	$3\times\left({4\over 3}\right)^2$	${\sqrt 3\over 4}\left[1+3\left({1\over 9}\right)+12\left({1\over 9}\right)^2\right]$

Now I made the following observations:

The length of each edge at the $(n+1)$th stage will be one third of the length of the edgeat the $n$th stage.
The number of edges at the ($n+1)$th stage is $4$ times greater then the number of edges at the $n$th stage.
From the last two observations the length of the curve is increased by a factor ${4\over 3}$ at each stage.
The area of each additional small triangle included inside the curve at the $(n+1)$th stage will be one ninth of the area of the additional triangles included at the $n$th stage.
After stage $2$ the extra area included within the curve at the $(n+1)$th stage is ${4\over 9}$ths of the extra area included at the $n$th stage.

Stage	Number of edges	Length	Area contained inside the curve
$n$	$3\times 4^n$	$3\times\left({4\over 3}\right)^n$	${\sqrt 3\over 4}\left [1+3\left({1\over 9}\right) \left(1 + \left({4\over 9}\right) + \left({4\over 9}\right)^2 + \ldots + \left({4\over 9}\right)^{n-1} \right)\right]$
$n \to \infty$	$\infty$	$\infty$	${2\sqrt 3\over 5}$

To find the area contained inside the fractal curve we need the infinite sum of the geometric series: $$1 + \left({4\over 9}\right) + \left({4\over 9}\right)^2 + \left({4\over 9}\right)^3 + \ldots$$ which is $${1\over \left(1-{4\over 9}\right)} = {9\over 5}.$$ Hence the area inside the fractal curve is ${2\sqrt 3\over 5}$.

In order to calculate the dimension $d$ in 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 $n =4$ and $m = 3$.

So, the dimension of the Von Koch Curve is $$d = {\log 4\over \log 3}= \log_3 4 \approx 1.262$$.