That's funny, I saw this problem before recently, and someone posted an answer of 2 + (3/2)?, which is the distance of a path wherein one walks straight out 1 mile, thence walks 3/4 of a circle, and then walks one more mile parallel to the first mile. That seemed like the right answer, but there is a shorter way. One should walk 2/?3 of a mile, thence another 1/?3 mile back at a angle of 60° to the first segment, then around 7/12 of a circle, and then 1 more mile tangent to the end of the circle, for a total distance of 1 + ?3 + (7/6)? which is just a bit shorter, about 6.39724, instead of 6.71239.
Can anyone beat this?
Edit: One way to find this angle of 60° is to work out the length as a function of the angle the first segment makes with the road that the last segment is perpendicular to, and find the minimum length. A briefer way might be to observe that 1) any straight line segment is always a tangent of the circle segment that it meets with, and 2) the 1st and 2nd segments makes the same angle with that road at equal angles, like a reflected light ray. Armed with 1) and 2), it's not hard to show that the angle has to be 60°. Maybe I should post a graphic?