Non-standard geometry problem

Solution 1:

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.

Solution 2:

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 $$A=\int\nolimits_K\ {\rm d}(x,y)\leq\int\nolimits_K\nu(x,y)\ {\rm d}(x,y)={1\over2}\int_0^{2\pi}r^2(\phi)\ d\phi\leq{\pi\over4}\ .$$