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.
A: I think you have to find an upper bound for the perimeter of the figure by using the fact that all splitting lines' length $\leq 1$ (I didn't yet, but it somehow seems obvious that there is one). Then you can use the Isoperimetric Inequality .
Hope that helps.
A: The area $A$ of the figure $K$ is in fact $\leq{\pi\over4}$, and this bound is attained only if $K$ is a circular disk of diameter $1$.
Let $\bigl(u(\phi),v(\phi)\bigr)$ be the midpoint of the median with slope $\phi$, and let  $\mu$  be the locus of these midpoints. Then the boundary curve $\gamma$ of $K$ has a parametric representation of the following form:
$$\left.\eqalign{x(\phi)&=u(\phi)+r(\phi)\cos\phi\cr
y(\phi)&=v(\phi)+r(\phi)\sin\phi\cr}\right\}\qquad(0\leq\phi\leq 2\pi)\ ,$$
where $u(\cdot)$, $v(\cdot)$ and $r(\cdot)$ are periodic with period $\pi$ (and not $2\pi$ !).
Denoting by $A_\phi$ the part of $A$ to the right of the median with slope $\phi$, $\ 0<\phi<\pi$, and by $A(\phi)$ its area, we have
$$2A(\phi)=\int_{\partial A_\phi}(x\ dy-y\ dx)=\int_{\phi-\pi}^\phi(x y'-yx')\ dt +\int_\sigma (x\ dy -y\ dx)\ ,$$
where $\sigma$ denotes the directed segment from $\bigl(x(\phi),y(\phi)\bigr)$ to $\bigl(x(\phi-\pi),y(\phi-\pi)\bigr)$. Using that $u$, $v$ and $r$ have period $\pi$ one computes
$$A'(\phi)=2 r(v'\cos\phi-u'\sin\phi)\ .$$
As this should vanish identically we necessarily have 
$$v'(\phi)\cos\phi-u'(\phi)\sin\phi\equiv0\ .\qquad\qquad(1)$$
Geometrically this means that  where $(u',v')\ne(0,0)$ the midpoint locus $\mu$ is the envelope of the medians.
Maybe there is a simpler way to prove that.
Let $R$ be the rectangle $[0,1]\times[0,2\pi]$ in the $(t,\phi)$-plane and consider the map
$$g:\ R\to K\ ,\quad (t,\phi)\mapsto\cases{x:=u(\phi)+tr(\phi)\cos\phi \cr y:=v(\phi)+t r(\phi)\sin\phi \cr}\quad.$$
This map is surjective, since through each point $(x,y)\in K$ there is at least one median, so that the forward half of a median turning around $360^\circ$ will pass over this point. Therefore the function $\nu(x,y)$ counting the inverse images of the point $(x,y)$ is $\geq1$ on $K$. The Jacobian of $g$ computes to
$$J_g(t,\phi)=r(v'\cos\phi-u'\sin\phi) + tr^2=t\> r^2(\phi)\geq 0\ ,$$
where we have used $(1)$. Using the (intuitively evident) formula
$$\int\nolimits_K\nu(x,y)\ {\rm d}(x,y)=\int\nolimits _R J_g(t,\phi)\ {\rm d}(t,\phi)$$ from geometric measure theory it now follows that
$$A=\int\nolimits_K\ {\rm d}(x,y)\leq\int\nolimits_K\nu(x,y)\ {\rm d}(x,y)={1\over2}\int_0^{2\pi}r^2(\phi)\ d\phi\leq{\pi\over4}\ .$$
