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Given a uniform planar convex region C, let us consider 2 centers of mass - the center of mass of the region as a whole and the center of mass of its boundary alone (assuming its boundary to have constant linear density). It is obvious that for all regular polygonal regions, both centers coincide. Even for other shapes such as the rectangle or ellipse or an equilateral triangle with all 3 vertices chopped off equally (it is now a hexagon), the two centers coincide.

Question: Are there uniform planar convex regions which have no rotational or reflection symmetry at all but with both centers of mass coincident? If so, how can we characterize regions where both centers of mass are coincident?

Note: This post continues Polyhedrons and their centers of mass. Based on this earlier post, one can think of other centers of mass and also 3D analogs to this question.

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In Euclidean triangle geometry the Spieker point $S$ is the centroid of the triangle sides, and it is known to be the anticomplement of the incircle center $I$ (https://faculty.evansville.edu/ck6/encyclopedia/ETC.html), which means it is in line with $I$ and the triangle centroid $C$, such that $$\overline{S C}~=~\frac{1}{2} \,\overline{I C}.$$ The following theorem generalizes this fact to $n$ polyhedra with inscribed spheres by a proof that is based on rescaling.

Definition:

The Spieker point of a $n$-polyhedron is the centroid of all its $n-1$ faces.

Theorem:

In every $n$-polyhedron $P$ with inscribed sphere, the incenter $I$, the centroid $C$, and the Spieker point $S$ lie in this order on a line, such that $$\overline{S C}~=~\frac{1}{n} \,\overline{I C}.$$

Proof:

Denote by $v$ the $n$-volume, and by $a$ the area of the $n-1$-faces, and by $r$ the radius of the inscribed sphere of $P$. For $\epsilon>0$, expand $P$ by a factor $1+\epsilon/r$ and translate it to $P_\epsilon$ such that the insphere centers of $P$ and $P_\epsilon$ coincide. Because the insphere radii are perpendicular to the faces in the touching points, the corresponding faces of $P$ and $P_\epsilon$ are parallel and have distance $\epsilon$. The volume of $P_\epsilon\backslash P$ is $ a_\epsilon ~=~\epsilon\, a +O(\epsilon^2)$, its centroid is $S_\epsilon$ and converges to $S$.
By the expansion around $I$, the centroid $C_\epsilon$ of $P_\epsilon$ lies on a line with $I$ and $C$ and
$$\overline{I C_\epsilon}~=~(1+\epsilon/r) \,\overline{I C}, $$ such that $$\overline{C C_\epsilon}~=~\epsilon/r \,\overline{I C}. $$ On the other hand, $C_\epsilon$ is the centroid of $C$ with weight $v$ and $S_\epsilon$ with weight $ a_\epsilon$, whereby $C,C_\epsilon,$ and $S_\epsilon$ lie on a line which then also contains $I$ and $$v\,\overline{C C_\epsilon}~=~\frac{v\,\epsilon}{r}\,\overline{I C}~=~a_\epsilon\, \overline{S_\epsilon C_\epsilon}. $$ Taking the limit $\epsilon \to 0$
results in $$\overline{S C}~=~\frac{v}{a\,r} \,\overline{I C}.$$ Observing that the volume of the $n$-pyramid over any $n-1$ face $F$ of $P$ with tip $I$ is $$ v(pyramid) ~=~\frac1n\,r\,area(F),$$ and that the sum $\frac1n\,a\,r$ of all these volumes is equal to the total volume $v$ of $P$, it follows that $$\overline{S C}~=~\frac{v}{a\,r} \,\overline{I C}~=~\frac{1}{n} \,\overline{I C},$$ which proves the theorem.

A corollary of this theorem is that in a $n$-polyhedron $P$ with inscribed sphere with centroid $\vec{c}$ and Spieker point $\vec{s}$ the incenter is given by $$\vec{i} ~=~ (n+1)\, \vec{c} - n\, \vec{s}.$$

As a consequence, centroid and Spieker point coincide in $n$-polyhedra with inscribed sphere if and only if incenter and centroid agree.

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    $\begingroup$ Thanks! Could you clarify if in the proof, the maximal inscribed sphere needs to touch all faces of the polyhedron? Iow, is the proof, by any chance, restricted to those polyhedrons wherein the max inscribed sphere touches all faces? $\endgroup$ Commented Jul 28, 2023 at 4:26
  • $\begingroup$ Thanks for the question. The proof uses the fact that by isotropic expansion around $I$ the faces move by a constant length, so touching is not reqired, but the incenter should be on the inside of all faces, i.e. in the halfspace in which the negative outer normal vector points. $\endgroup$ Commented Jul 28, 2023 at 10:13
  • $\begingroup$ But note that the proof also uses that $P$ is star-shaped with respect to $I$. $\endgroup$ Commented Jul 28, 2023 at 10:20
  • $\begingroup$ Thanks again. Guess one can construct a convex planar region with both I and C coincident and without any symmetry by taking an equilateral triangle T and cutting off three small pieces with mutually different shapes from T's corners such that the resulting body T' is convex and also has same C as T. If the cut pieces are small, the I of T' will obviously be that of T. From your proof, T' will have both CMs coincident. Here, T' will have at least 6 sides. So finding a region with less sides (3 might not be possible) and with both CMs coincident might be more interesting. $\endgroup$ Commented Jul 28, 2023 at 11:33

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