can a convex polygon have only one boundary point at locally maximum distance from its centroid?

Question

It's easy to see that given any convex polygon P and any point c in its interior, there is at least one point m on the boundary of P at locally maximum distance from c: simply choose m to be a vertex at maximum distance from c.

The following picture shows an example of P and c such that P has only one boundary point m at locally maximum distance from c.

Which of the following statements are true? I'm looking for a simple proof or counterexample for each.

(S₀₂) Every convex polygon P has at least two boundary points at locally maximum distance from the centroid c₀ of its vertices.

(S₁₂) Every convex polygon P has at least two boundary points at locally maximum distance from the centroid c₁ of its boundary.

(S₂₂) Every convex polygon P has at least two boundary points at locally maximum distance from the centroid c₂ of its area.

These are all cases of the more general parametrized statement, for $0 \leq$ k $\leq$ n:

(S_kn) Every convex n-dimensional polytope P has at least two boundary points at locally maximum distance from the centroid c_k of its k-skeleton.

Answer 1

Here are the the answers to some of the cases. S₂₂ is the one in which I was most interested.

• S₀₀: false by inspection.
• S₀₁: true by inspection.
• S₁₁: true by inspection.
• S₀₂: false (extra points can be introduced to skew $c_0$)
• S₁₂: true, I think (sketch of possible proof below)
• S₂₂: true (proof below)
• S₀₃: false (extra points can be introduced to skew $c_0$)
• S₁₃: false (extra edges can be introduced to skew $c_1$)
• S₂₃: ?
• S₃₃: false (not too hard to construct a counterexample; note that Unistable Polyhedra are counterexamples to the dual statement)

Proof of (S₂₂) Every convex polygon P has at least two boundary points at locally maximum distance from its area centroid:

Given any polygon $P$ and an interior point $c$, let "local maxima" and "local minima" denote those boundary points which are at locally maximum and minimum distance from $c$. Since the boundary of a polygon follows no circular arcs, the local maxima and minima are well defined and isolated, and there are only a finite number of each. Since every continuous function on a compact set achieves its minimum, compactness of the boundary of $P$ implies that, when the boundary is traversed in a counterclockwise direction, between every pair of local maxima there is a local minimum, and similarly between every pair of local minima there is a local maximum. Therefore local minima and local maxima alternate, and there is an equal number of each.

Now, assume we are given $P,c$ such that $P$ has only one local maximum with respect to $c$; we must show that the area centroid of $P$ is not $c$. From the previous paragraph, since there is exactly one local maximum, there is also exactly one local minimum. Call the local minimum $b^-$ and the local maximum $b^+$, at distances $r^-$ and $r^+$ from $c$.

When the boundary is traversed in a counterclockwise direction, the part from $b^-$ to $b^+$ has strictly increasing distance from $c$, and the part from $b^+$ to $b^-$ has strictly decreasing distance from $c$. Therefore by the intermediate value theorem, for any $r$ strictly between $r^-$ and $r^+$, there will be two boundary points at that distance from $c$, one in the increasing part and one in the decreasing part.

So if we construct a circle centered at $c$ of radius $r^-$ at time $t=0$, and continuously grow it to radius $r^+$ at time $t=1$, then at all intermediate times, the circle will intersect the boundary of $P$ in exactly two places; the two intersection points start together at $b^-$, then they separate, one moving around the circle clockwise and the other counterclockwise, until they finally meet at $b^+$. By another application of the intermediate value theorem (applied to the continuously increasing angle between the rays from $c$ to the two intersection points), there will be some time $t$ such that the two intersection points are directly opposite each other on the circle, so that $c$ is collinear with them. Let $C$ be the circle at that time $t$. See diagram.

Let $L$ be the line passing through $c$ and the two points of intersection of $C$ with the boundary of $P$; $L$ cuts $C$ exactly in half. Let $P^+$,$P^-$,$C^+$,$C^-$ be the parts of $P$ and $C$ on the $b^+$ and $b^-$ sides of $L$. Then $P^- \subset C^-$, and $C^+ \subset P^+$. The containments are strict, in the sense that there is nonempty area left over in each case.

Therefore when computing the area moment of $P$ about $c$, $P^-$ is exactly balanced by the 180-degrees-rotated-around-c copy of $P^-$ which is strictly contained in $C^+ \subset P^+$. The remainder of $P$ has nonempty area and is entirely on the $b^+$ side of $L$; therefore its centroid also lies on that side of $L$, and therefore isn't $c$. So the overall area centroid of $P$, being a nontrivial weighted average of $c$ and some point that isn't $c$, also isn't $c$. Q.E.D.

Sketch of possible proof of (S₁₂):

Most of the proof is the same as that of S₂₂, up through $P^- \subset C^-$ and $C^+ \subset P^+$. The only part that is not entirely clear is that the boundary of $P^+$ must then "outweigh" the boundary of $P^-$, putting the centroid off-balance from $c$. The length of $P^+$ is certainly greater than that of $P^-$, by the lemma proposed in this question . So the desired proof might require an argument similar to the one used in the accepted answer to that question.