can a convex polygon have only one boundary point at locally maximum distance from its centroid? 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:
(Skn) Every convex n-dimensional polytope P has at least two boundary points at locally maximum distance from the centroid ck of its k-skeleton.
 A: 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.
