geometry - Blindfolded and disoriented near the Great Wall of China

You are blindfolded and disoriented, standing exactly 1 mile from the Great Wall of China. How far must you walk to find the wall?

Assume the earth is flat and the Great Wall is infinitely long and straight.

Answer

$\DeclareMathOperator{\arcsec}{arcsec}$

For each possible orientation of the wall (relative to some arbitrary initial orientation), the point on the wall closest to our starting point is a distance $1$ away. The collection of the closest points for all possible orientations of the wall form a circle of radius $1$ around our starting point.

If we move a distance $r>1$ away from the initial point, we intersect two orientations of the wall that are an angle $\theta$ apart. In order to reach that point we must have crossed all of the orientations in that angle. In the figure below on the left, those "explored" points are marked by a magenta line.

By trigonometry we can show that $\theta = 2\arcsec r$. If we traverse the path shown on the right side of the above figure, we travel a worst-case distance of:

$$ r + r(2\pi - \theta) \\ r + 2r(\pi - \arcsec r) $$

This distance is minimized when $r\approx 1.04356$ for a worst-case distance of $6.99528$, an improvement of about $3.95\%$

However, looking at the figure we can immediately see that the majority of the large circular arc is "wasted" distance. Only the ends contribute to additional "explored" points. If we shrink-wrap the rest of the path around the unit circle, we get the following path:

The worst-case distance of this path is:

$$ r + 2\sqrt{r^2-1} + (2\pi - 2\theta) \\ r + 2\left(\sqrt{r^2-1} + \pi - 2\arcsec r\right) $$

This happens to be minimized for $r = \sqrt{\frac{15-\sqrt{33}}{6}} \approx 1.24200$ (not the distance shown in the figure), for a worst-case distance of:

$$ \sqrt{\frac{9+\sqrt{33}}{2}}+4\arctan \sqrt{\frac{9+\sqrt{33}}{8}} \approx 6.45891 $$

an improvement of $11.32\%$.

Update

Thanks to Michael Seifert for pointing out that we can do better by letting the radii of the start and end be different, in which case we have the distance:

$$ r_1 + \sqrt{r_1^2-1} + \sqrt{r_2^2-1} + 2\pi - \theta_1 - \theta_2 \\ r_1 + \sqrt{r_1^2-1} + \sqrt{r_2^2-1} + 2\pi - \arcsec r_1 - \arcsec r_2 $$

Which is minimized by $r_1=2/\sqrt{3},\ r_2=\sqrt{2}$ (with $\theta_1=\pi/3,\ \theta_2=\pi/2$):

(Because of the nice angles, this picture is exactly to scale.) The worst-case distance here is simply

$$ \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} + \frac{2\pi}{3} + \frac{\pi}{2} + 1 \\ = 1 + \sqrt{3} + \frac{7\pi}{6} $$

(a $12.16\%$ improvement.)