Let $P$ be a simple, planar polygon. Define $c_a$ as the area centroid of $P$, i.e., the center of gravity of the closed shape $P$. Define $c_p$ as the perimeter centroid of $P$, the center of gravity of just the boundary edges of $P$. One can compute $c_p$ as the weighted sum of the edge midpoints, weighted by their length, and normalized by the total perimeter of $P$.

My question is:

Q. Is there a clean characterization of those $P$ for which $c_a = c_p$?

Of course there are obvious examples (e.g., the regular polygons), but there are also non-obvious examples. Here is one I constructed:

$$P= ((-1, 3), (-0.0426, 0), (-1, -0.0213), (0, -0.0426), (1, -0.0213), (0.0426, 0), (1, 3), (0, 1)) \;,\; \implies c_a =c_p = (0,1.247) $$
It seems these polygons are rare but not as uncommon as one might at first think. It would be interesting to quantify "rare" and "not as uncommon."

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    $\begingroup$ with triangles, the area centroid equals the vertex centroid, but usually not the edge centroid. That surprised me at some point. $\endgroup$ – Will Jagy Dec 30 '14 at 3:17
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    $\begingroup$ You should be able to take any two polygons (rotated so their four centroid points are all collinear) and glue appropriately scaled versions together with a thin corridor so that the resulting polygon has centroids that are coincident. $\endgroup$ – Michael Biro Jan 1 '15 at 23:49
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    $\begingroup$ Posted similar (but not the same) question [math.stackexchange.com/questions/1173903/… There is another fundamental centroid: mass center of vertices. Are there any polygons, except regular, with all three centers coincide? $\endgroup$ – lesobrod Mar 3 '15 at 18:51
  • $\begingroup$ @lesobrod: That should be easy by sprinkling extra vertices collinear with edges in my example. These extra vertices will not change area or perimeter centroids, but will change the vertex centroid. $\endgroup$ – Joseph O'Rourke Mar 3 '15 at 19:00

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