## Area bounded by the Koch Anti-Snowflake Curve

The first observation is that the area of a general equilateral triangle
with side length **a** is

\[\frac{1}{2} \cdot a \cdot \frac{{\sqrt 3 }}{2}a = \frac{{\sqrt 3 }}{4}{a^2}\]

as we can determine from the following picture

For our construction, let the length of the side of the initial triangle be given by the value of *s*. By the result above, using **a** = *s*, the area of the initial triangle s therefore
\(\frac{{\sqrt 3 }}{4}{s^2}\).

To form the Koch anti-snowflake curve, start with an equilateral triangle and begin the iteration steps for constructing a copy of the Koch curve along
each of the three sides, but pointing *inside* the original equilateral triangle.

The areas of the newly formed equilateral triangles must be subtracted. So at the first iteration (middle figure above) we have 3 sides and along each side we remove a triangle with side length *s*/3. The area bounded by the curve at this iteration is therefore

\[\frac{{\sqrt 3 }}{4}{s^2} - 3 \cdot \frac{{\sqrt 3 }}{4}{\left( {\frac{s}{3}} \right)^2} = \frac{{\sqrt 3 }}{4}{s^2}\left( {1 - \frac{3}{9}} \right)\]

We have 4 new edges from each of the 3 original sides, for a total of 12 edges. Therefore at the next iteration (right figure above) we remove the areas of 12 new triangles each of which have a side of length *s*/9. At this iteration the area of the bounded region is

\[\frac{{\sqrt 3 }}{4}{s^2}\left( {1 - \frac{3}{9}} \right) - 3 \cdot 4 \cdot \frac{{\sqrt 3 }}{4}{\left( {\frac{s}{9}} \right)^2} = \frac{{\sqrt 3 }}{4}{s^2}\left( {1 - \frac{3}{9} - \frac{{3 \cdot 4}}{{{9^2}}}} \right)\]

Continuing with the construction, at the kth iteration we remove 3×4^{k-1} additional triangles of area \(\frac{{\sqrt 3 }}{4}{\left( {\frac{s}{{{3^k}}}} \right)^2}\). The total area that must be subtracted is

\[3 \cdot {4^{k - 1}} \cdot \frac{{\sqrt 3 }}{4}{\left( {\frac{s}{{{3^k}}}} \right)^2} = \frac{{\sqrt 3 }}{4}{s^2}\left( {\frac{{3 \cdot {4^{k - 1}}}}{{{9^k}}}} \right)\]

Hence after n iterations the area of the bounded region is

\[\frac{{\sqrt 3 }}{4}{s^2}\left( {1 - \sum\limits_{k = 1}^n {\frac{{3 \cdot {4^{k - 1}}}}{{{9^k}}}} } \right)\]

The sum inside the parentheses is the partial sum of a geometric
series with ratio **r** = 4/9. Therefore the sum converges as
**n** goes to infinity, so we see that the area bounded by the Koch anti-snowflake curve is

\[\begin{align}
\frac{{\sqrt 3 }}{4}{s^2}\left( {1 - \sum\limits_{k = 1}^\infty {\frac{{3 \cdot {4^{k - 1}}}}{{{9^k}}}} } \right) &= \frac{{\sqrt 3 }}{4}{s^2}\left( {1 - \frac{{3/9}}{{1 - 4/9}}} \right) \\
&= \frac{{\sqrt 3 }}{4}{s^2}\left( {\frac{2}{5}} \right) \\
\end{align}\]

Since \(\frac{{\sqrt 3 }}{4}{s^2}\) is the area of the original equilateral triangle, this shows that the bounded region has area equal to 2/5 that of the original triangle.