Challenge Level

Jeremy of Drexel University and Andrei of Tudor Vianu National College, Bucharest, Romania sent in excellent solutions. The first half of this solution is Andrei's, the second Jeremy's.

To solve this problem, I made a table with the length and the area of the fractal at the first stages:

Stage | Length | Area |

0 | 4 | 1 |

1 | $4\times2=8$ | 1 |

2 | $4\times2^2=16$ | 1 |

I have made the following observations:
The length of the fractal at each stage is twice its length at
the preceding stage.
The area of the fractal is constant because the two colored
areas are equal (the blue one is added, the yellow one
subtracted)
For the $n^{th}$ stage the length of the curve is $4 \times 2^n =2^{n+2}$ and the area is $1$ square unit |

Now we have the curve length for any stage $n$, and in turn, a stage $n$ for any particular curve length. \begin{eqnarray} 1,000,000 &=& 2^{n + 2}\\ n &=& log_2 (1,000,000) - 2 \\ n &\approx& 17.9 \end{eqnarray} Because 17 < n < 18 , the length must have been less than $1,000,000$ at stage $17$ and greater than $1,000,000$ at stage $18$. Thus answers the question: stage $18$. Further, we can now find the exact lengths of the curve at stage $17$ and stage $18$ to convince ourselves: $$L(17) = 2^{17 + 2} = 2^{19} = 524,288$$ which is less than $1,000,000$ $$L(18) = 2^{18 + 2} = 2^{20} = 1,048,576$$ which is greater than $1,000,000$.

The length of the fractal is $$\lim_{n\to \infty} 2^{n+2}$$ and hence the fractal curve has infinite length. This is an example of a curve of infinite length surrounding a finite area.

What is the dimension of the fractal? Solution: $d = 1.5$

Explanation: the dimension $d$ is given by the formula $n = m^d$.

$n = 8$ because each segment is broken up into 8 self-similar segments.

$m = 4$ because each of the $8$ new segments is $1/4$ the length of the original segment, and thus must be multiplied by $4$ to be the length of the original segment.

So: $8 = 4^d$ and hence $d = log_4 8 = 1.5$.