Lost in Illinois

The problem can be restated as: "Find the minimum length $s$ of the path S with convex hull of $\frac{120}{\pi} m^2$ " (the trick here is to realize that we need not to, and also should not, return to the starting point), as we do know that the cornfield is convex. Now we must prove that the path S should be a half-circumference (so the convex hull becomes a semi circle).
The method is to compare any other path S' with that S (same convex hull area, of course): Just reflect the convex hulls of S and S' over the line connecting the start point and the end point, we will get new shapes P and P' with the same area of twice the area (of that cornfield). Using the fact that the shortest curve to enclose an area is the circle, which is now P, any P' with the same area has bigger perimeter, meaning longer path S'. (QED)
Demonstration
So, with $r$ and $A$ as the radius and the area of the semi circle,
$s^2 = (\frac{2\pi r}{2})^2 = 2A \times \pi = 2 \times \frac{120}{\pi} \times \pi = 240$

Consider the semicircle with radius $\frac {\sqrt{240} }{\pi}$ , and curved length $\sqrt{240}$ . This semicircle has area $\frac {120} {\pi}$ . Let's walk along this path. If we don't walk out of the cornfield, then the cornfield must enclose an area of $\frac{ 120 } { \sqrt{\pi} }$ , since it contains all of these line segments.

Assume that it is possible to exit by walking a smaller path of length $L < \sqrt{240}$ . Reflect this path along the line connecting the 2 endpoints, and we get a cycle of length $2L$ that encloses an area of $2 \times \frac {120}{\pi}$ , which contradicts the given fact.

Hence, $D = \sqrt{240}$ , so $D^2 = 240$ .

If you did not know it already, the problem tells you (in the Details and Assumptions section) that a Circle encloses the most Area per the length of its perimeter. Thus, intuition would tell you to simply walk a circle, finding its radius from the given Area of the field, yielding an answer of 480. However, this is not optimized.

This is where the fact that the field is comes in. Since the field is convex, walking only half of a circle would still enclose the area of a semi-circle because your beginning point and endpoint would connect to each other without you having to physically walk that distance. While it may seem that this is pointless because you would walk half of the curve to get half of the area, since the area of a circle increases quadratically with radius and the Circumference of a circle increases linearly with radius, when we scale up the circle such that the semi-circle encloses an area of \frac{120}{π} , our total length walked will be smaller.

From this information, we can finally set up our equation.

\frac {1}{2} πr^2 = \frac {120}{π} \rightarrow r = \frac {4\sqrt{15}}{π}.

To find the circumference of the curvature of the semi-circle we would have to walk we simply multiply by π. This leaves us with

s = 4\sqrt{15} \Rightarrow s^2 = 240

Thus, the answer is 240.

Note: Using this method with even smaller partitions of circles such as a quartercircle does not work as the starting and end points of your path are not concurrent with the center of the circle, leaving you with significantly less area.

The smallest path that can be formed is the circumference of a semicircle that has an area of the convex object

so if $s$ is the circumference of the semicircle of radius $r$ , then $s=r \pi$ . by dividing both sides by $\pi$ , we will have $r=\frac{s}{\pi}$

In the formula of the area of a semicircle $\frac{1}{2}\pi r^2 = \frac{120}{\pi}$ , we substitute the radius we got from the circumference which is $r=\frac{s}{\pi}$

by that we will have $s^2 = 240$

It turns out that the optimal solution is when the path is the arc of a semicircle (of area $A = \frac{120}{\pi}$ ), which is easily manipulated using area and circumference formulae to obtain $s^2 = 240$ - the details are less interesting than proving that the solution is a semicircle.

Now, to show that the optimal path is a semicircle:

Note that the worst case scenario is when the entire path is inside the convex field. Thus, the path must have a convex hull of area at least $A$ .

Suppose that we start the path at P and end at Q (where PQ runs East-West WLOG). We want to find the optimal solution for given P,Q - in fact, we want to show that this is the arc of a circle. South of the line PQ, draw the arc of the circle which is the same circle that contains an area A above the line. That is, there is a unique circle through P and Q, where the North section, along with PQ, bounds area A - draw the southern section. Suppose that the southern section bounds an area B with PQ.

Now, the shortest curve which bounds an area A+B is a circle congruent to the one above (by the quotable result). That means that the shortest curve which bounds an area A with the line PQ must indeed be the arc of the circle, as required.

Now, note that we can ignore the case P=Q since that gives a circle, with circumference $\sqrt(480)$ , which is greater than we had above. Finally, if we relax the area restriction but try to determine the minimum value of $\frac{s^2}{a}$ where s denotes arclength and a denotes bounded area, we note that this is the same problem since we can scale the solution to this to get the correct answer, and it is easy to determine by differentiating that the ratio is minimised when we indeed have a semicircle.