Chebyshev center = center of mass?

I would like to know for which convex polyhedra $P$ in $\mathbb{R}^3$, is the center of the largest sphere enclosed in $P$ (a.k.a. the Chebyshev center, or the incenter) the same as the center of mass/gravity of $P$, assuming $P$ is a homogenous solid. The conditions force quite a bit of symmetry. Perhaps only the Platonic and Archimedean solids have the property.

The question can be asked in all dimensions, and I do not even know the answer for $\mathbb{R}^2$.

I feel this question must have been explored, but I have not found it my literature searches. I'd appreciate any pointers. Thanks!

Edit. Here are the examples suggested by Rahul and Eric. The left image is from Wikipedia's page on bipyramids. The right image I made myself.

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Apart from the Platonic and Archimedean solids, I think the isohedral bipyramids also have this property. –  Rahul Mar 23 '11 at 19:31
Would not a prism formed from a rotationally symmetric polygon work as well provided it was just long enough to make the incenter unique? –  Eric Nitardy Mar 23 '11 at 21:10
@Rahul and Eric: I think you are both correct! Nice. The rotational symmetry ensures the conditions are satisfied. Cool. Now the challenge is to determine the full class of such polyhedra. But this is a great start. Thanks! –  Joseph O'Rourke Mar 23 '11 at 23:45
It'd be nice to know the answer in the plane anyway. –  lhf Mar 24 '11 at 0:04
@lhf: It seems as though there will be lots of diverse and sundry polygons that work. Take a circle and three distinct points on that circle. Imagine three sides of a polygon lying on the tangent lines for the three points. Now connect those three side by inserting any number of sides between them. You need not worry about symmetry. Just make sure that the weight of the sections of the polygon between the three points of tangency balance so that the centroid is the center of the circle. I'd be more hopeful of an interesting result if the circumcenter and centroid were coincident. –  Eric Nitardy Mar 24 '11 at 6:16

Too long for a comment:

There are other obvious shapes, such as suitable anti-prisms.

But I don't think you are going to find any symmetry is a requirement, at least if you have a large number of faces. You could take something with a 100 faces which you have already found works, and regard finding the in-centre as an optimisation problem subject to the constraints of being the correct side of each face. Because you have so many faces acting as constraints, you can relax a number of them without allowing the in-centre to move. So stick asymmetric pyramids or frusta or other things on several of these relaxed faces so the new solid is still convex and the centroid remains at the in-centre.

Or you could cut off several corners of the original solid in asymmetric ways, without affecting the in-sphere or moving the centroid. Once you have this, you probably don't even need symmetry of the points where the in-sphere is tangent to the faces of the solid, since you can introduce new tangents by cutting off corners suitably and then remove old tangents by stick other bits on. Then all symmetry is lost.

I expect the same applies in $\mathbb{R}^2$. Indeed I suspect the minimal number of edges of an asymmetric convex polygon with the in-centre coinciding with the centroid is 4, 5 or 6.

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Your reasoning is compelling (and a bit disappointing). Thanks for enlightening me. –  Joseph O'Rourke Mar 24 '11 at 1:33
Re. "Too long for a comment": I think this is perfectly valid as an answer, as it does address the original question ("Which convex polyhedra have this property?") in the negative sense ("Too many to describe neatly."). –  Rahul Mar 24 '11 at 3:45