Lost in the Woods

In this picture 10 units = $1000$ meters.

Basically, you need to ensure you hit every tangent to an imaginary circle. You begin at the center, O, and this circle has radius $1000m$ .

Pick a reference direction and call it "north". Turn 30 degrees to the east of north and walk $\frac{2000}{\sqrt{3}}\approx 1154.70m$ to point A.

Next turn 120 degrees to the right and walk $\frac{1000}{\sqrt{3}}\approx 577.35m$ to point B. You are now 1000m from the start but at an angle 60 degrees east of north.

Next walk $\frac{7}{12}$ of the way around this imaginary circle of radius $1000m$ to point C.

Finally, walk $1000m$ due north to point D.

Total distance is $1000\sqrt{3} + \frac{7000\pi}{6} + 1000 \approx \boxed{6397}m$

There seems to be a lot of interest in proving this is the optimal solution. I'm not sure I could prove it. J. R. Isbell did find and prove this path is optimal in 1957. The best I can do is feel proud that I was able to find it independently.

I can share my thought process (as I recall it from a couple of years ago)

Clearly, one way to guarantee to find the road is to go 1000 meters in some direction (it may as well be called north) and then walk a full circle.

Ok, can I do better? Think about the road as possible tangents. Well, at the end of the route instead of the last quarter circle you could go due north. This saves you some walking as $\pi/2<1$ . This is segment CD.

But we can't do that at the start. Well, we sort of can if we go exactly North-East for $1000\sqrt{2}$ . Then due south for 1000 to hit the circle. This saves a bit because $\sqrt{2}<\pi/2$ . These portions were then refined.

What if we went up at some other angle, maybe more North than East as sort of a balance between the two other ideas. Pick a point A somewhere at the y=1000 line, walk to A and then back to the circle at B. I'm pretty sure I used calculus to minimize the two segments and the small bit of arc.

That brought me to the solution I found. I tried also making the curved path to C shorter or longer. I also tried making D higher or lower in hopes of adjusting A, but this didn't help either. When I reported my findings to the person who showed the problem to me, they indicated it was the solution.

This result was initially proved by J.R.Isbell here in 1957.

Just to add to the solutions provided, the first step in the chosen route identifies an angle, x and traces a route similar to the ones shown of 1/cos (x) away from the starting location then sin (x)/cos (x) back to the circle.

The distance saved traversing the circle is 2x which leaves a simple case of minimising the route which we do by differentiating the equation:

(1+sin(x))/cos(x) -2x

Working through the maths this gives 1 + sin x = 2 cos(x)^2,

=> x=30°

Using Jeremy's figure, let's say you go in some direction until you think you've gone far enough, and then turn back, thus possibly just missing a road, which is DA. Then if you travel on a circle of radius 1000 meters, you will touch all the places where the road can be until you are at C and headed back in the direction of road DA. CD is the shortest path to that road. So, the problem remains to be solved is the shortest distance from O to A to B to any point on the circle of radius 1000 meters. For shortest path, the angle OA mades with line DA is the same with the angle BA makes with line DA, and line AB should be tangent to the circle of radius 1000.

Given any line DA tangent to the circle, if one goes from O to any point on line DA, thence around the circle, and then back any point on line DA, this is the shortest posssible path.

1st you need to walk in a direction 1000m, then turn 90degrees in a direction left or right to start walking on the edge of a hypothetical circle, thus the distance traveld to ensure u reach the road is antything between 1000 and 1000 plus 2000 x 3.14, as for calculating a deffinate shortest distance, this problem is flaud