Q:

Non-standard geometry problem

I am trying to solve the following problem:

A plane, convex, bounded figure has the property that any chord which splits it in half las length at most $1$. Prove that the figure has area less than $2$.

I'm not sure how to approach it, and how to use the fact that the splitting line has length less than 1. I guess there exists an approach with integrals, but I try to find a more elementary solution, since this is from an olympiad problem book. Thank you.

A:

Here is a sketch of a geometric solution that gives an area bound of 2. For simplicity, assume that the boundary has a unique tangent at every point.
Now I claim that there exists a halving chord $c$ such that the tangents at its two endpoints are parallel. The simplest way to see this is to start with an arbitrary halving chord and draw tangents at its endpoints. Continuously move it so that it always stays a halving chord. The chord must return to its original position in the opposite orientation. During this time, the angle between the tangents must go from $\theta$ to $-\theta$ so the tangents must be parallel at some point during this process.
Now the figure is sandwiched between two lines at unit distance. Since every halving chord intersects $c$, every point on the figure is at most distance 1 from $c$. Thus, the figure lies inside a $1\times 2$ rectangle.

