# Number of bitangents to convex polytopes

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):

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 $$P_2$$:

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 $$d \ge 3$$?

• 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. – Gerhard Paseman May 15 at 15:32
• @GerhardPaseman: Tried to make it more clear with another image... – Joseph O'Rourke May 15 at 16:36
• 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. – Gerhard Paseman May 15 at 16:49
• By the way, the bitangents should be (d-1)-dimensional, not d dimensional. Gerhard "Exponential Error Easily Escalates Excitement" Paseman, 2019.05.15. – Gerhard Paseman May 15 at 17:23
• 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. – Gerhard Paseman May 15 at 21:52

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.
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.