# 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.

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How are you going to use convexity and boundedness? –  Mark Bennet Aug 12 '11 at 6:22
@anon: A chord which splits the figure in half need not pass through the centroid. Consider dividing an equilateral triangle in half by a chord parallel to one of the sides. –  Rahul Aug 12 '11 at 7:16
Since you mention the integral approach you might be familiar with this: i1.informatik.uni-bonn.de/publications/gkms-chapc-07.pdf . If you were not, now you are. –  Kai Aug 12 '11 at 11:36
@Kai: Thank you very much for the article. –  Beni Bogosel Aug 12 '11 at 14:10
The presented article clearly gives a solution, since somewhere in it there is an inequality between area and the square of the greatest halving chord.(I guess there is a small error in the article in the respective inequality since the circle does not verify the given inequality there.) The solution uses integrals and the area is smaller than $\pi/4$ which is really smaller than 2... I would be glad if there would be a elementary solution. Anyway, the constant 2 is way to big so that there must be a geometric method. –  Beni Bogosel Aug 12 '11 at 15:02

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.

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Thanks. I think this works even if the boundary doesn't have a unique tangent. –  Beni Bogosel Aug 13 '11 at 5:37

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

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Very nice. I didn't have time to check the details because I'm at work, but thank you for your answer. –  Beni Bogosel Aug 13 '11 at 12:15
The difficult part of your argument looks suspiciously close to one proof of the isoperimetric inequality. Maybe you can simplify that using lemma 1.11.1 on p.21 of Struwe's DG-script and estimating the area as in (1.11.1) on p.20 there? I didn't look that closely, though. –  t.b. Aug 14 '11 at 19:24

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.

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