A closed figure with an infinitely long perimeter ... what?

One of the classic questions in the field of fractal geometry is: How long is the coast of Britain? At first, it seems like there's an easy answer: Just take a ruler, walk around the island, and measure it.

But the coast of Britain is quite rocky and convoluted. What if you used a shorter ruler, say a centimeter (cm) in length instead of a meter (m), and measured around the biggest of the rocks. You'd get a longer result. And if you used a shorter ruler, the result would be longer still. Imagine using a ruler so small that you could measure around the bumps of each atom that made up each rock and grain of sand along the coast ... longer still.

This central conundrum from fractal geometry can be modeled using geometric figures like the Koch snowflake, a closed figure (with an inside and an outside, like a circle) that can be manipulated so that it maintains its symmetry and area, but has an unlimited number of sides and an infinite perimeter.

How it works

Look at the sequence of figures below. We start with an equilateral triangle (left). Now remove the middle third of each side, and in its place, build in another equilateral triangle. The length of each side of the new triangular protrusion is 1/3 the length of the previous side.

The new closed figure then has 12 sides of equal length, and four times as many sides as the original triangle. Notice that the area of the second figure is larger than that of the initial triangle, but in later steps, the area will grow more slowly, ultimately reaching a limit.

We can repeat this side-generating process as many times as we want. Here are the first three iterations. You can see how the number of sides (N), the length of each side (S) and the perimeter (P sum of the lengths of all sides) changes in the table below.

Now let's try to analyze what's going on here. First the number of sides seems to change by a factor of four on each iteration, from 3 to 12 (4 × 3) to 48 (4 × 12), and so on. If we call the number of sides of each figure N, and let No be the number of sides of the starting triangle, N1 be the number of sides of the first iteration, and so on, here's how we calculate the number of sides.

Now if we notice the pattern – 4 to some power – we can find a formula for the number of sides of any iteration:

$$N_n = 3 \cdot 4^n$$

OK, not how about the length of each side? If we call the length of a side of the original triangle s, then we notice that the length of each side after each iteration will be 1/3 of the previous side length:

$$ \begin{align} S_o &= s \\ S_1 &= \frac{s}{3} \\ S_2 &= \frac{s}{3} \cdot \frac{1}{3} = \frac{2}{9} \\ S_3 &= \frac{s}{9} \cdot \frac{1}{3} = \frac{s}{27} \end{align}$$

... and we can also find a formula for the length of a side Sn, for any iteration:

$$S_n = \frac{S_{n - 1}}{3} \, ,$$

where Sn-1 is the length of the side of the previous iteration, Sn is the current iteration, and n is just a counter of how many times we've added sides. To put it another way,

$$S_n = \frac{s}{3^n} \, ,$$

because 32 = 9, 33 = 27, and so on.

The perimeter grows without bound

Finally, we can calculate the perimeter. It's just the length of a side (Sn) multiplied by how many sides there are (Nn):

$$P_n = N_n \cdot S_n = (3\cdot 4^n) \left( \frac{S}{3^n} \right)$$

Then a quick rearrangement gives us a formula that's enlightening:

Now the important thing to notice is that the length of the perimeter of a Koch Snowflake has no bound. The more sides we add (the higher n gets), the larger the perimeter. So we have a closed figure with an area that is bounded (I haven't proved that here), but a perimeter that can be infinitely long. That's weird.

The perimeter is infinite, but the enclosed area is not ...

Now let's use a similar approach to calculate the total area of the Koch snowflake. If we divide the starting triangle into smaller triangles, the side of each being a third of the side of the original, we get nine triangles, like this:

Now we form the first iteration by attaching three of these to the original triangle, like this:

The area of one of those new triangles is 1/9 of the original, bringing the total area to

$$A = a_o + 3 \frac{a_o}{9} = a_o\left( 1 + \frac{3}{9} \right)$$

where ao is the area of the original triangle; presumably this is something we'd know at the beginning.

The next iteration involves adding a set of smaller triangles, each a ninth of the small red triangle above, and therefore each 1/92 or 1/81 of the original triangle:

We add four of those new triangles to each of the 3 sides of the original figure for a total of 4·3 = 12 new triangles, and the growing area is now

$$A = a_o \left( 1 + \frac{3}{9} + 3 \cdot 4 \cdot \left(\frac{1}{9}\right)^2 \right)$$

Let's do one more iteration and look for a pattern. The next set of triangles to be added will have 1/93 the area of the original triangle, ao, and there will be 16·3 of them, like this:

The new area expression then looks like this:

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$$A = a_o \left[ 1 + 3\cdot \frac{1}{9} + 3 \cdot 4 \cdot \left( \frac{1}{9}\right)^2 + 3 \cdot 16 \cdot \left( \frac{1}{9} \right)^3 \right]$$

Now it's pretty easy to see a pattern emerge. The trick is to reduce it to some form, probably a series, that we can analyze for convergence (heading for some finite limit) or divergence (growing infinitely large). The pattern looks like this:

$$A = a_o \left[ 1 + 3\cdot \frac{1}{9} + 3 \cdot 4 \cdot \left( \frac{1}{9}\right)^2 + 3 \cdot 16 \cdot \left( \frac{1}{9} \right)^3 + \dots + 3(4^{n - 1}) \left( \frac{1}{9} \right)^n \right]$$

Notice that there are powers of 4 and 9 in there, but that the powers of 4 lag behind those of 9. If we factor out a 1/9, we can make them the same and simplify this series a lot:

$$A = a_o \left[ 1 + \frac{3}{9}\left( 1 + \left( \frac{4}{9} \right)^1 + \left( \frac{4}{9} \right)^2 \right) + \left( \frac{4}{9} \right)^3 \right]$$

Cleaning up and adding some summation notation, we get:

$$A= a_o \left[ 1 + \frac{3}{9} \sum_{n = 0}^{\infty} \left( \frac{4}{9}\right)^n \right]$$

Now this embedded series (the sum) is a convergent geometric series because the part raised to the power n is less than one. I leave it to you to calculate the actual value of this sum, but what's really important is that the area reaches some finite limit while the perimeter does not.

That's what makes the Koch snowflake and figures like it so interesting, and a good model for the length of a coastline. Thinking about figures like this has led to development of the whole field of fractal geometry, which I hope you'll explore more on your own.

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