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Let me state my question prior to defining terms:

Q. Let $P_1$ and $P_2$ be disjoint convex polytopes in $\mathbb{R}^d$ of $n$ vertices each. What is the maximum number of distinct bitangent $(d{-}1)$-dimensional hyperplanes realizable, as a function of $n$ and $d$, the maximum over all such polytopes?

In $\mathbb{R}^2$, the hyperplanes are lines, and the polytopes are polygons.

Say a hyperplane $H$ is tangent to a polytope $P$ if (a) $H$ contains at least one vertex, and (b) the interior of $P$ lies to one side of $H$. $H$ is a bitangent to $P_1$ and $P_2$ if it is (a) tangent to both (and so contains at least one vertex of both), and (b) $H$ contains at least $d$ vertices total. Say that two bitangents are distinct if the vertices of the polytopes they include are not identical.

So in $\mathbb{R}^2$, a bitangent contains $\ge 2$ vertices, and the answer to Q is $4$ independent of $n$ (thanks to Gerhard Paseman for this):


          SqTangents
          Bitangents to squares.


In $\mathbb{R}^3$, a bitangent plane includes $\ge 3$ vertices. One can arrange two polyhedral convex cones $P_1$ and $P_2$ so that for each of $n-1$ vertices of $P_1$, there are $n-1$ different bitangent planes through two vertices of $P_2$:


          Cones
          Bitangents to polyhedral cones. Two bitangents shown.
So the answer to Q for $d=3$ is $\Omega(n^2)$. Correction. As Gerhard pointed out, many of these supposed bitangent planes cut the cones. So in fact this example only shows $\Omega(n)$ bitangent planes.

My question is: How many distinct bitangent hyperplanes can there be in dimension $d$, for $d \ge 3$?

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  • $\begingroup$ I challenge your assertion that for each point there are n-1 bitangent planes in your cones example. I can believe there are $\Omega(n)$ for n-1 points, but a proof would be nice. Gerhard "Not All Planes Are Bitangent" Paseman, 2019.05.15. $\endgroup$ – Gerhard Paseman May 15 at 15:32
  • $\begingroup$ @GerhardPaseman: Tried to make it more clear with another image... $\endgroup$ – Joseph O'Rourke May 15 at 16:36
  • $\begingroup$ Thank you for the image. If the statement is that there are n-1 triangles (or planes), (not bitangents), then I agree. Otherwise, you are saying that for every edge in the lower n-1-gon, there is a bitangent that includes that edge and any one of the n-1 vertices of the upper n-1-gon. I do not believe that. Gerhard "I'm Still Not Bi-ing It" Paseman, 2019.05.15. $\endgroup$ – Gerhard Paseman May 15 at 16:49
  • $\begingroup$ By the way, the bitangents should be (d-1)-dimensional, not d dimensional. Gerhard "Exponential Error Easily Escalates Excitement" Paseman, 2019.05.15. $\endgroup$ – Gerhard Paseman May 15 at 17:23
  • $\begingroup$ When I extend the triangle containing vertices 2 and 3, I get it cutting the other cone. Even if point 1 is external and admits many bit an gent a to the other edges, each edge can only support two bitangents. Gerhard "An Edge Looks One Co-dimensional" Paseman, 2019.05.15. $\endgroup$ – Gerhard Paseman May 15 at 21:52
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Note that a plane containing $k \lt d$ points of a convex polytope must contain the face having those $k$ points. So a weak upper bound on your number is a product of two numbers, each of which counts the number of permissible faces of each polytope.

Let's suppose we have fixed d-1 points in general position of our prospective bitangent and are seeking another point to fix it in place. If the first d-1 are distributed among both polytopes, then the next candidate (a) has to be in general position with the first d-1 points, and (b) has to be adjacent as in belonging to a polytope face that contains some of the d-1 points. So this limits the possibilities significantly.

If one polytope has all d-1 points, then there is even less freedom. As in the two dimensional case, sweeping a plane around a fixed subspace of codimension 1 gives up to four distinct possibilities for bitangential contact with at least one of two convex bodies. So I challenge the lower bound assertion ($\Omega(n^2)$ at this writing) for the cone example.

Gerhard "Let's Face Up To It" Paseman, 2019.05.15.

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