Starting with a convex polyhedron $P_1 \subset \mathbb{R}^3$, replace that with $P_2$, the convex hull of the midpoints of the edges of $P_1$. Continuing this process, we obtain a series of polyhedra approaching a smooth body $B$ (or at least, I think it approaches a smooth body). See above for $P_1,\ldots,P_5$—not to the same scale.

Q1. Is $\lim_{n \to \infty} P_n$ $C^2$-smooth, starting with non-degenerate $P_1$? Or only $C^1$-smooth? Or only $C^0$?

Q2. Does $P_n$ approach an ellipsoid as $n \to \infty$, for every (non-degenerate) starting $P_1$? [My guess: No.]

These are, in some sense, less sophisticated versions of my earlier question, “Derived” polyhedra and polytopes, which focussed on face centroids rather than edge midpoints. But here I am asking specific questions on smoothness and the limit shapes. Still, perhaps @GjergjiZaimi's answer there holds.

The limit objects are almost subdivision surfaces, but not quite.

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    $\begingroup$ There's a point (is it the centroid?) on each face that also lies on some face of each iterated polyhedron. So if there is a limiting surface all those points lie on it. It seems to be easy to make a convex polytope so that there is no ellipsoid intersecting every face. $\endgroup$ – Brendan McKay Oct 20 '14 at 6:27
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    $\begingroup$ I conjecture the complete answer to Q1 and Q2 should be: $B$ is always $C^1$ smooth (that's should be easy, since a convex is $C^1$ iff any point has a unique leaning plane), but it is $C^2$ only if it is an Euclidean ball, and $P_1$ is a regular polyhedron. $\endgroup$ – Pietro Majer Oct 20 '14 at 10:27
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    $\begingroup$ But for the same reason $P_2, P_3,\dots$ should be all regular polyhedra, which is impossible, so I'd say as a conjecture that the limit convex in never $C^2$. $\endgroup$ – Pietro Majer Oct 21 '14 at 10:19

The limit can be within $\varepsilon$ of any convex polytope:

Start with a polytope $P$ then replace each vertex $v$ by three vertices $v_1,v_2,v_3$ forming a triangle with sides of length (less than) $\varepsilon$ with centroid $v.$ and containing no point of $P$ other than $v.$ This gives a polytope $P_1 \supset P$ with three times as many vertices but with all points within $\varepsilon$ of (the closest point of) $P.$ Then as the edge midpoint process is applied to $P_1$ one will get a nested series of polytopes all containing $P.$ $P_1$ will have many faces aside from the tingy triangles and the centroids of these faces will be in the limit body, so the limit os strictly larger than $P.$

I'd still like to know the exact limit for a regular tetrahedron (or irregular, it shouldn't matter) and other simple cases.

Here are the first nine stages for $P_1$ the cube (not shown) with vertices $(\pm 2,\pm 2,\pm 2).$ The centroids of the $6$ faces, at distance $\sqrt{4}$ from the origin, will be in the limit body. One is $[2,0,0].$

Then $P_2$ is a cuboctahedron ($14$ faces, $12$ vertices) introducing $8$ new centroids all at distance $\sqrt{16/3}$. One is $[4/3 ,4/3,4/3].$

Then $P_3$ has $26$ faces and $24$ vertices. This introduces $12$ centroids all at distance $\sqrt{9/2}.$ One is $[0,3/2,3/2].$

After this is $P_4$ with $50$ faces and $48$ vertices. The $24$ new centroids are all at distance $\sqrt{147/32}.$ One is $[7/8,7/8,7/4].$

Finally in this picture is $P_5$ with $98$ faces and $96$ vertices. There are $48$ new centroids introduced. Half are like $[0,7/8,15/8]$ at distance $\sqrt{137/32}$ and half are like $[7/8,23/16,23/16]$ at distance $\sqrt{627/128}.$

I'm not sure how much this reveals other than that the limit body is not a sphere. The centroids introduced, which will remain centroids at each stage and are extreme points of the limit body are $6$ at distance $2$ then $8$ at distance$\approx 2.3094$ then $12$ at distance $\approx 2.1213$ then $24$ at distances $\approx 2.1433$ and finally $24$ each at distances $\approx 2.0691,2.2132.$

enter image description here

In general it is non-trivial to find the faces and edges for the convex hull of a set of points (even given that all the points are vertices.) However given the information on $P_i$ it is easy to get $P_{i+1}$: Replace each face by the midpoint polygon and for each vertex find the faces it is on, from each one take an edge joining the midpoints of the two appropriate edges and then use these edges to list these points in a valid order.

Here are $P_i$ for $i=6,7,8,9.$ Each is shown twice. Once centered on a triangle and once not. Notice that the even and odd cases look different. In the top view of $P_8$ it may look as if three triangles and the diamonds meet at a degree $6$ point. However that point is actually a tiny triangle and the "triangles" are trapezoids with a very short side.

enter image description here

The "equator" for $P_9$ looks like this

enter image description here

  • $\begingroup$ Construction from the point 2 (bullet) $\ P\subset\cap_{n=1}^\infty P_n\subset P_1\ $ is a very nice idea. On the other hand I don't understand point 1. By taking all midpoints you're getting $\ P_1\ $ back, i.e. by iterating we get $\ P_1\ P_1\ P_1\ \ldots\ $-- and this is not a typo. I must be missing something. $\endgroup$ – Włodzimierz Holsztyński Oct 20 '14 at 15:56
  • $\begingroup$ I'm taking out that first point. It is not wrong in the way that you say, but it does change things. Consider what happens to a square face. $\endgroup$ – Aaron Meyerowitz Oct 21 '14 at 1:51
  • $\begingroup$ A tetrahedron produces an octahedron, and both the octahedron and the cube produce the cuboctahedron. So the limit convex is the same for these three regular polyhedra... $\endgroup$ – Pietro Majer Oct 21 '14 at 10:08
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    $\begingroup$ I think one can prove: If the limit convex is an Euclidean ball, then $P_1$ must be inscribed and circumscribed to concentric spheres . This implies that all its edges have the same length, and that its faces are regular polygons. So it must be an Archimedean Polyhedron. But the same conclusion should then hold for $P_2,P_3,\dots$ which is impossible because there are finitely many Archimedean polyhedra. So I think an Euclidean ball can never be realized as a limit convex. $\endgroup$ – Pietro Majer Oct 21 '14 at 10:27
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    $\begingroup$ The computation is quite simple given the starting polyhedron (as a set of ordered lists of points , but no concern about orientation.) I'm not even sure that the triangle points are that sharp. They may look so in the illustration because there are other points close by and edges converging from several directions. In a drawing without edges shown they might not stand out as much. If , on $P_n$, you go on the surface from the center of a triangle to the center of one of the three neighbor triangles then you hit many vertices or none depending on the parity of $n$ $\endgroup$ – Aaron Meyerowitz Oct 22 '14 at 1:54

As Wlodzimierz mentioned, if a plane contains a face of $P_1$, it contains a face of every $P_n$. Now consider a finite set of planes $W_i$ (I think $6$ will do) such that no ellipsoid is tangent to all of them but some convex body has faces in all of them. Then the limiting shape can't be an ellipsoid. For example, this should be the case if the initial polyhedron is a prism with the base pictured here:

enter image description here

  • $\begingroup$ A convincing answer to Q2---Thanks! $\endgroup$ – Joseph O'Rourke Oct 20 '14 at 11:10

I don't have a full answer. I hope that my observation provides some insight and a starting point. Let all vertices have weight $\ 1.\ $ Consider an arbitrary plane $\ W\ $ which contains face $\ F\ $ of $\ P_n\ $ (in particular of $\ P_1).\ $ Then plane $\ W\ $ contains the respective faces of all next generation polyhedrons, and the center of vertices of $\ F,\ $ and of all next generation faces contained in $\ W\ $ is the same point belonging to the respective polyhedrons $\ P_{n+d}\ (d=0\ 1\ \ldots)$. Thus all the mentioned centers belong to the topological boundary of intersection $\ \bigcap_{n=1\ 2\ \ldots} P_n$.

Furthermore (an EDIT session), each such center is the only point of the respective plane $\ W.\ $ This is a very strong indication that the intersection $\ \bigcap_{n=1\ 2\ \ldots} P_n\ $ is strictly convex.


A variant of Robert Israel's argument: Take a convex body $K$. Slice it by planes, to make a polyhedron $K_\epsilon\subset K$ whose Hausdorff distance to $K$ is less than $\epsilon^2$ and the diameter of faces is less than $C\epsilon$. Apply the edge-midpoint process. Then the limit body has many points at distance $<\epsilon^2$ to $\partial K$; one per face. Thus the set of limit bodies is dense in the set of convex bodies. In particular, most limit bodies are not ellipsoids.

Alternately, you can form a polyhedron $K^\epsilon$ containing $K$, arbitrarily close to $K$, by taking finitely many tangent planes. Then again apply the edge-midpoint process to obtain a limit body close to $K$.


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