Let me state my question prior to defining terms:

. 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?Q

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

**for $d=3$ is $\Omega(n^2)$.**

*Q***. As Gerhard pointed out, many of these supposed bitangent planes cut the cones. So in fact this example only shows $\Omega(n)$ bitangent planes.**

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