Let $C$ be a convex polygon in the plane and let $s$ be the shortest line segment (I believe this is called a "chord") that divides the area of $C$ in half. What is the smallest angle that $s$ could make where it touches the boundary of $C$? This picture shows an example, I I called the angle $\theta$ (and the area of the region is $A$):
Numerical simulations suggest that $\theta\geq\pi/3$ but I haven't been able to prove this.
$\let\eps\varepsilon$This is not a complete answer. I will just show that $\theta$ can be smaller than $\pi/3$, but $\theta>\pi/4$.
1. Take an isosceles triangle $XYZ$ with $\angle Y=\angle Z=\pi/3+2\eps$ for a sufficiently small $\eps$. Cut out a small isosceles triangle $XY'Z'$. The resulting trapezoid $YZZ'Y'$ is a counterexample if $\eps$ and $XY'$ are chosen appropriately.
Namely, if $\eps$ and $XY'$ are small enough (the latter needs not be very small), then there are four halving chords with locally minimal length, joining $Y'Y$ with $YZ$, $Z'Z$ with $YZ$ (these two are equal), $Y'Z'$ with $YZ$ (this one is much larger if $XY'$ is small), and $YY'$ with $ZZ'$ (this one is larger than the first ones if $XY'$ is not too small comparable with $\eps$). Thus the first two are the shortest ones, and they subtend $\pi/3-\eps$.
Right now I cannot say how far this example can be extended. Also, it seems to be almost obvious that it is better to make $Y'Z'$ some curve rather than a line segment.
2. Now assume that the shortest halving segment of length $a$ joins $X$ and $Y$ and subtends equal angles $\theta<\pi/4$ at both its endpoints $X$ and $Y$. (It is easy to see that the shortest halving segment should connect two points different from vertices, and the subtended angles are equal). The supporting lines at $X$ and $Y$, together with $XY$, bound a triangle $\Delta$ of area $<a^2/4$, and a piece of $\partial C$ on its side of $XY$ lies inside $\Delta$.
Consider a ray perpendicular to $XY$, and move its endpoint along this piece of $\partial C$; at some moment it will contain a segment $I$ halving the area. The length of $I$ is at least $a$, the part of $I$ inside $\Delta$ is less than $a/2$, so its part outside $\Delta$ is larger than $a/2$. But then the convex hull of this part, together with $XY$, already makes an area $>a^2/4$, and lies in $C$. This contradicts the assumption that $XY$ is area-halving.
This proof works also for $\theta=\pi/4$, as the only polygon for which we get equalities everywhere is a square, but then it contains a halving segment with less length.
Again, this proof can be improved, but I do not know right now how far.
I'll prove that for any triangle below, $\displaystyle \theta \geqq \pi /3\ $
Assuming $\displaystyle X'Y' =z$ be the line segment that bisects area $\displaystyle A$ of triangle $\displaystyle XYZ$ .
Let $\displaystyle \angle Z=\alpha $ , $\displaystyle Y'Z=x$ and $\displaystyle X'Z=y$
By Heron's formula, $\displaystyle \frac{A}{2} =\frac{1}{4}\sqrt{( x+y+z)( x+y-z)( x-y+z)( -x+y+z)} =\frac{1}{4}\sqrt{( 2xy)^{2} -\left( x^{2} +y^{2} -z^{2}\right)^{2}}$ ,
so we get $\displaystyle z=\sqrt{x^{2} +y^{2} -2\sqrt{( xy)^{2} -A^{2}}} \geqq \sqrt{2xy-2\sqrt{( xy)^{2} -A^{2}}}$ .
On the other hand, $\displaystyle \frac{A}{2} =\frac{1}{2} xy\sin \alpha $, so $\displaystyle xy=\frac{A}{\sin \alpha }$,
and we get $ $$\displaystyle z\geqq \sqrt{2A\left(\frac{1}{\sin \alpha } -\frac{1}{\tan \alpha }\right)} =\sqrt{2A\tan\frac{\alpha }{2}}$
When $\displaystyle x=y$ and $\displaystyle \alpha $ be the smallest angle of triangle $\displaystyle XYZ$, $\displaystyle z$ get the minimum value.
and then $\displaystyle \theta =\frac{\pi -\alpha }{2}$, so $\displaystyle \theta \geqq \pi /3\ $.
