TITLE: Can this ant find its way back to the nest?
QUESTION [36 upvotes]: So the puzzle is like this:

An ant is out from its nest searching for food. It travels in a straight line from its nest. After this ant gets 40 ft away from the nest, suddenly a rain starts to pour and washes away all its scent trail. This ant has the strength of traveling 280 ft more then it will starve to death. Suppose this ant's nest is a huge wall and this ant can travel in a whatever curve it wants, how can this ant find its way back? 

I interpret it as: I start at the origin. I know that there is a straight line with distance 40 ft to the origin, but I don't know the direction. In what parametric curve I will sure hit the line when the parameter $t$ is increasing, while the total arc length is less than or equal to 280 ft.
I asked a friend of mine who is a PhD in math, he told me this is a calculus of variation problem. I wonder if I could use basic calculus stuff to solve this puzzle (I have learned ODE as well). My hunch tells me that a spiral should be used as the path, yet I am not sure what kind of spiral to use here. Any hint shall be appreciated. Thanks dudes!
Clarification by dfeuer
As some people seem to be having trouble understanding the problem description, I'll add an equivalent one that should be clear:
Starting in the center of a circle of radius 40 ft, draw a path with the shortest possible length that intersects every line that is tangent to the circle.

REPLY [4 votes]: This is only a partial proof and it is intended only as an illustration of my comment to @copper.hat - a full proof seems already be given the other answer...   
I'm not sure with the second part: The second part is, that we assume , the minimal path ends with a path walking on the circumference of the circle and finally on the vertical line down to the horizontal "wall-shaped nest"; this means the upper half circumference ( $p_4=r \cdot \pi$ ) and one radius ( $p_5 = r \cdot 1$ ) are assumed to be the minimal trailing path after the leading part has been done and thus is fixed to the length $p_4+p_5 = r \cdot (\pi+1)$  where $r=40$ is the radius.  
The initial part of the path is then the variable part of the problem.
The ant has to go down to the wall with some angle $\varphi$ which is the first part of the path ( $=p_1$ ) , then it has to go back to the circumference ( second part $p_2$ ) , and walk on the circumference to the position $\pi/2$ on the circumference which is the end of the first quadrant (third part $p_3$ ).     

Obviously that part on the circumference is $p_3=r\cdot (\pi/2-2\varphi)$        
The part $p_1$ depends on the angle $\varphi$: $p_1 = r \cdot {1 \over\cos(\varphi)}$ and     
the part $p_2$ is $ p_2 = p_1 \cdot \sin(\varphi)$

So th variable part of the ant's path is $f(\varphi) = p_1 + p_2 + p_3 = r\cdot \left({1+\sin{\varphi}\over \cos\varphi} + \pi/2 - 2 \varphi \right) $ and for this a minimizing expression can be given: $ {d \over d \varphi} f(\varphi) = r \cdot \left({\sin^2 \varphi+\sin \varphi\over \cos^2 \varphi}-1\right) $   and the solution occurs indeed as $\varphi = 2 \pi/12 = 30°$ which is the half angle of the hexagon.     
However, this is not yet the full proof; perhaps there is a somehow fractal path crossing the circumference to the inside of the circle, do some shortcut there and proceed - it must be shown, that there is indeed no shorter path for this part.