A claim on the concurrency of area bisectors of planar convex regions

Question

We add a little bit to On 'fair bisectors' of planar convex regions and Bisectors and partitioning lines for convex regions defined with respect to the moment of inertia

Definitions: Given a planar convex region C (could be smooth or polygonal), an area bisector of C is any line that partitions C into 2 pieces of equal area. A perimeter bisector is a line that partitions C into 2 pieces of equal perimeter. Obviously, thru every point on the boundary of C we can draw an area bisector and a perimeter bisector.

Question: Are the following claims easy to prove/counter?

• A planar convex region is centrally symmetric if and only if its area bisectors are all concurrent.
• A planar convex region is centrally symmetric if and only if its perimeter bisectors are all concurrent.

Note: Higher dimensional analogs of these claims are easy to state.

Answer 1

• A planar convex region is centrally symmetric if and only if its area bisectors are all concurrent.

Yes. Let all area bisectors of our region $K$ pass through a point $O$. Note that any line $\ell$ through $O$ is an area bisector of $K$: otherwise there exists a parallel area bisector (by continuity), and it does not pass through $O$. Assume that $K$ is not $O$-symmetric. It yields that some chord $AB$ through $O$ is not bisected by $O$. Move it slightly to get a chord $A_1B_1$ ($A_1$ close to $A$). The curvilinear triangles $AOA_1$ and $BOB_1$ have unequal area, thus at least one of chords $AB$, $A_1B_1$ is not an area bisector.

Answer 2

At least if we also assume that the region is $C^1$, then the second claim is also true.

Parametrise the region's boundary with polar coordinates $r=r(\theta)$ where the origin is the intersection of all its perimeter bisectors. Note that the length of the perimeter curve for $a\leq \theta\leq b$ is $\int_a^b \sqrt{r(\theta)^2+r'(\theta)^2}d\theta$, therefore the claim that every line through the origin is a perimeter bisector translates to $$ r(\theta)^2+r'(\theta)^2 = r(\theta+\pi)^2 + r'(\theta+\pi)^2 $$ The result is a conclusion of the following claim:

Claim. Let $f, g$ be nonnegative $C^1$ functions, periodic with a common period. If we have $f'^2+f^2 = g'^2+g^2$ identically, then $f=g$.

Note that nonnegativity is necessary, otherwise the functions $f=1, g=\cos\theta$ are counterexamples. The requirement of $C^1$ is also necessary - $f,g$ being merely continuous, and $C^1$ outside a finite set is not enough by the counterexample $f=1, g=\max(\cos(\theta), \cos(\theta+\frac23\pi), \cos(\theta+\frac43\pi))$ (however this counterexample gives rise to a nonconvex region, I am not sure if there is a convex counterexample).

To prove the claim, assume to the contrary that there is some $x_0$ for which $f(x_0)\neq g(x_0)$. Without loss of generality we may assume that $f(x_0) < g(x_0)$. Note that $f'(x_0)^2 - g'(x_0)^2 = g(x_0)^2 - f(x_0)^2 > 0$. In particular, $f'(x_0)\neq 0$, and if we assume wlog that $f'(x_0)>0$, then we get that $\left|g'(x_0)\right|<f'(x_0)$ and therefore $g-f$ is positive decreasing near $x_0$.

Let $I$ be the largest open interval containing $x_0$ such that $f<g$ and $f'>0$ on $I$. Then, as above, we get that $g-f$ is positive decreasing in $I$. Let $x_1=\inf I < x_0$ (note that $I$ is bounded because $f$ is periodic). Then on one hand (from $g-f$ being decreasing) we get that $g(x_1) - f(x_1) > g(x_0) - f(x_0) > 0$, and on the other hand (from definition of $I$) $x_1$ being an endpoint of $I$ means that either $f(x_1)=g(x_1)$ or $f'(x_1)=0$, and note that $f'(x_1)=0$ also implies $f(x_1)=g(x_1)$: $$ f(x_1)^2=f'(x_1)^2+f(x_1)^2 = g'(x_1)^2+g(x_1)^2 \geq g(x_1)^2 \geq f(x_1)^2 $$ and we get a contradiction.