Slaying the Dragon

Let $d$ be the straight-line distance from one end of the road to the other. The fractal may be constructed by an iterative process. At the beginning, the middle third of a straight line with length $d$ is removed and replaced with two segments of the same length as the removed segment, adding a total length of $\frac{d}{3}$ . Then, each of the two added segments with length $\frac{d}{3}$ have a third of their length added in the same way, adding a total of $\frac{2d}{9}$ . After that, each of the four segments added in the previous step have a third of their length added, for a total of $\frac{4d}{27}$ , and so on. The final length of the fractal is

$d\left(1 + \frac{1}{3} + \frac{2}{3^2} + \frac{2^2}{3^3} + \frac{2^3}{3^4} \dots \right) = d\left(1 + \frac{1}{3}\left(1 + \frac{2}{3} + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 \dots\right)\right)$ .

The geometric series $1 + \frac{2}{3} + \left(\frac{2}{3}\right)^2 ...$ is equal to $\frac{1}{1 - 2/3} = 3$ , so the expression evaluates to $d\left(1 + \frac{1}{3} \cdot 3\right) = 2d$ . Therefore, if $d =$ 15 leagues, the total distance along the road is 30 leagues and the knight will travel for $\frac{30 \: \text{leagues}}{9 \: \text{leagues/hour}} = \frac{10}{3} \: \text{hours}$ in lowest terms. 10 + 3 = 13 .