1. Start with an equilateral triangle.
2. Divide each side into three equal parts.
3. One the middle third of each side draw an equilateral triangle.
4. Delete the 'base' of the triangle to produce one inside area.
5. Go back to step 2 and repeat.
The resulting curve is shown below.
The first iteration produces a shape with 12 sides, each of length
\[\frac{1}{3}\]
, so the perimeter is \[12 \times \frac{1}{3}=4\]
.The perimeter is \[12 \times \frac{1}{3}=3 ^1 \times (\frac{4}{3})^1\]
The second iteration produces a shape with 48 sides, each of length
\[\frac{1}{3^2}\]
, so the perimeter is \[48 \times \frac{1}{3^2}=3 \times (\frac{4}{3})^2\]
.And so on.
The shape has
\[3 \times 4^{m-1}\]
sides, each of length \[\frac{1}{3^{n-1}}\]
, so the perimeter is \[3 \times (\frac{4}{3})^{n-1}{3})^2\]
.As
\[n \rightarrow \infty\]
, so does the perimeter.If the area of the original triangle is
\[A\]
then the area of the 2nd shape is \[A(1+ 3 \times \frac{1}{3^2})\]
.The area of the 2nd shape is
\[A(1+ 3 \times \frac{1}{3^2})\]
.The area of the 3rd shape is
\[A(1+ 3 \times \frac{1}{3^2})+12 \times (\frac{1}{3})^3)\]
.The area of the nth shape is
\[A(1+ 3 \times \frac{1}{3^2})+12 \times (\frac{1}{3})^3)+ ...+ 3\frac{4^{n-2}}{3^{2n-2}})\]
.This last expressiuon tends to a limit as
\[n \rightarrow \infty\]
.