- Start with an equilateral triangle.
- Remove the middle third of each side.
- Add two line segments to make new equilateral triangles on each side of the previous triangle.

*It is continuous everywhere (there are no breaks or discontinuities) but it's differentiable nowhere*. That means you can't calculate the slope at any point on it. Think about it! Try to imagine drawing a tangent line to determine the slope from first principles....you just can't. The line wiggly and bumpy no matter how much you magnify the shape. It's just not smooth anywhere along it.*The total area of the snowflake is finite but the perimeter is infinite*. The mathematical proof of this is given below. But the fact it fits on your computer screen or smartphone shows it has a finite area. You can see that the perimeter increases at each step by looking at the pictures. But not all increasing sequences have infinite sums. The series \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \) also increase but doesn't get bigger than 2. The proof that the perimeter of the snowflake is infinite is also given below.

## The finite area of the Koch Snowflake

Consider a triangle with three equal sides of length \(a\).

We can calculate the area using the familiar \(\frac{1}{2}\times base \times height\) formula.

In this picture we would write the area as \(\frac{1}{2}ah\). The height is found using the definition of the sine function and is \(a\sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{h}{a}\).

Therefore the triangle area is \[Area = \frac{\sqrt{3}}{4}a^2\]

Let's calculate these quantities.

First, the number of triangles added at each iteration.

This is a geometric series with a common ratio of 4 and first term 3. It means the number of triangles added at the \(k\)th iteration is \(3\times 4^{k-1}\).

\[\frac{\sqrt{3}}{4}\left(\frac{a}{3^k}\right)^2\]

\[3\times 4^{k-1} \times \frac{\sqrt{3}}{4}\times \frac{a^2}{3^2}\]

We can simplify this using the rules for indices to get

\[\frac{\sqrt{3}}{4}a^2\frac{3}{4}\left( \frac{4}{9} \right)^k\]

The total area of the Snowflake is the area of the first triangle plus the sum of areas for each iteration \(k=1,2,3\ldots\)

\[\displaystyle \textrm{Snowflake area} = \frac{\sqrt{3}}{4}a^2 + \sum_{k=1}^{\infty} \frac{\sqrt{3}}{4}a^2 \frac{3}{4}\left( \frac{4}{9} \right)^k\] Take the constant terms outside the summation:

\[\displaystyle \textrm{Snowflake area} = \frac{\sqrt{3}}{4}a^2 +\frac{\sqrt{3}}{4}a^2 \frac{3}{4} \sum_{k=1}^{\infty} \left( \frac{4}{9} \right)^k\]

The summation term is a geometric series with common ratio \(\frac{4}{9}\) and first term \(\frac{4}{9}\) so we can use the "sum to infinity" formula to evaluate it.

\[\displaystyle \textrm{Snowflake area} = \frac{\sqrt{3}}{4}a^2 +\frac{\sqrt{3}}{4}a^2 \frac{3}{4} \left( \frac{\frac{4}{9}}{1-\frac{4}{9}} \right)\]

Which simplifies to

\[\displaystyle \textrm{Snowflake area} = \frac{\sqrt{3}}{4}a^2 +\frac{\sqrt{3}}{4}a^2 \frac{3}{4} \left( \frac{4}{5} \right)\]

Take a common factor:

\[\displaystyle \textrm{Snowflake area} = \frac{\sqrt{3}}{4}a^2 \left( 1+\frac{3}{4}\frac{4}{5} \right)\]

And simplify to

\[\displaystyle \textrm{Snowflake area} = \frac{\sqrt{3}}{4}a^2 \left( \frac{8}{5}\right)\]

The Snowflake area is finite and \(\frac{8}{5}\) times bigger than the first triangle.

## The infinite perimeter

Therefore the perimeter is simply the number of sides \(\times\) side-length.

We already have a formula for the number of sides - we can modify slightly to give \(3\times 4^k\) sides on the \(k\)th iteration. Check it! It gives 3,12,48,... starting with \(k=0,1,2,3...\)

The side length at the \(k\)th iteration is \(\frac{a}{3^k}\)

The perimeter after \(k\) iterations looks like this:

\[\textrm{Snowflake perimeter} = 3\times 4^k \times \left(\frac{a}{3^k}\right)\]

which can be rearranged to give

\[\textrm{Snowflake perimeter} = 3a \left(\frac{4}{3}\right)^k\]

As \(k\) tends towards larger and larger values then perimeter increases without limit.

If the initial triangle has sides 10cm long then after 5 iterations the perimeter has increased to 1.26m. It rapidly increases. After just 73 iterations the perimeter is longer than the mean Earth-Moon distance! As the iterations increase the perimeter just keeps getting ever longer.

The Koch Snowflake has an infinitely long perimeter enclosing a finite area.