Blindfolded and disoriented near the Great Wall of China

Question

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 1

$\DeclareMathOperator{\arcsec}{arcsec}$

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.)

Answer 2

If the angle between the possible wall and the initial line is $x$ (the angle between the diagonal line and the bottom line in the diagram below), then the distance travelled is $1+(\pi/2+2x)+1/tan(x)+1/sin(x)$.

Gratifyingly this gives a slightly improved answer of $2+3\pi/2\approx6.7124$ for my first attempt (because you can drop down straight rather than complete the circle), where $x=\pi/2$.

It also gives my second attempt for $x=\pi/4$ (answer $2+\sqrt{2}+\pi\approx6.5558$).

Throwing the expression into wolfram alpha, shows that a minimum occurs at $\pi/3$. This gives a value of $1+\sqrt{3}+7\pi/6\approx6.397$

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Old new upper bound: $2+\sqrt{2}+\pi$ as per diagram:

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(Old upper bound: $2\pi+1$ miles. Walk 1 mile in any direction and then walk is a circle of radius 1, centred at your starting point. )

Answer 3

I would like to present this non-rigorous but hopefully more intuitive explanation for the optimal path. (The technique used here was very helpful for working on Oray's variant with two people.)

The first part of 2012rcampion's answer explains that we should go as far out as some tangent $l$, before going around the circle to get back to $l$ on the other side. Call the starting point $A$ and the circle $O$. Then the problem is this:

Find the shortest path that comes from $A$, touches $l$, then goes around the circle and touches $l$ again.

It won't change which path is shortest if we turn around at the end and go all the way back:

Find the shortest path that comes from $A$, touches $l$, then goes around the circle and touches $l$ again, and then goes back around the circle to $l$ and then $A$.

Now, if we reflect the entire diagram over $l$, we get this:

Instead of having our path touch $l$ and go back, we can have it switch sides every time instead, which won't change the length because it's just a reflection. So now the problem is this:

Find the shortest path from point $A$ that goes around circle $O^\prime$, then around circle $O$, then goes to point $A^\prime$.

Anyone should be able to do that (imagine putting a string from $A$ around the circles to $A^\prime$ and pulling it tight):

And now if we only look at the part of the diagram above $l$, there's the answer without any calculations.

Answer 4

"...standing exactly 1 mile from the Great Wall of China. How far must you walk to find the wall?"

You must walk 1 mile. If you go the wrong way then you will end up walking further. If you don't walk that far you can't reach it.

Answer 5

@DrXorile is clos to the answer. Mine isn't an answer either but here's some food for thoughts

I wanted to picture it. It looks like it.

If we take 360 individuals, all starting at the center of the circle and each at a different angle, only one will find the wall.

That's a 0.27% chance of finding the wall if you walk eaxtly one mile. If you need to reach the wall for your survival, you're dead.

Also imagine the guy who started just one degree slightly off, extend his hands and the wall is just 2 inches further, then starts over in the wrong direction.

Walking more than one mile means we could increase our chances of reaching the wall at slightly off angle.

But then again, this could happen: