Koch Snowflake - Properties

Properties

The Koch curve has an infinite length because each time the steps above are performed on each line segment of the figure there are four times as many line segments, the length of each being one-third the length of the segments in the previous stage. Hence, the total length increases by one third and thus the length at step n will be (4/3)n of the original triangle perimeter: the fractal dimension is log 4/log 3 ≈ 1.26186, greater than the dimension of a line (1) but less than Peano's space-filling curve (2).

The Koch curve is continuous everywhere but differentiable nowhere.

Taking s as the side length, the original triangle area is . The side length of each successive small triangle is 1/3 of those in the previous iteration; because the area of the added triangles is proportional to the square of its side length, the area of each triangle added in the nth step is 1/9 of that in the (n-1)th step. In each iteration after the first, 4 times as many triangles are added as in the previous iteration; because the first iteration adds 3 triangles then the nth iteration will add triangles. Combining these two formulae gives the iteration formula:

where is area of the original triangle. Substituting in

and expanding yields:

In the limit, as n goes to infinity, the limit of the sum of the powers of 4/9 is 4/5, so

So the area of a Koch snowflake is 8/5 of the area of the original triangle, or . Therefore the infinite perimeter of the Koch triangle encloses a finite area.

It is possible to tessellate the plane by copies of Koch snowflakes in two different sizes. However, such a tessellation is not possible using only snowflakes of the same size as each other. Since each Koch snowflake in the tesselation can be subdivided into seven smaller snowflakes of two different sizes, it is also possible to find tessellations that use more than two sizes at once.