Pushing a convex cone and equidistants

Question

Let $K$ be a closed convex cone in an n-dimensional Euclidean space. Suppose $K$ has non-empty interior. For $t > 0$ form the subcone $K_t$ consisting of all points in $K$ which lie a distance $t$ or greater from the boundary of $K$.

True or False? $K_t$ is a translate of $K$.

In other words, can we obtain $K_t$ by pushing $K$ into its interior?

Motivation: this problem arose while trying to establish the ‘optimal worst-case escape rate’ from the collision locus in the $N$-body problem. In the collinear equal mass $N$-body problem, the $N-1$ dimensional cone $K$ of interest is the one defined by the inequalities $ x_1 \le x_2 \le \dotsb \le x_N$ within the hyperplane $\sum x_i = 0$ of ${\mathbb R}^N$. (You might call this the “$A_N$ cone” since it is a Weyl chamber for the $A_N$ Lie algebra.)

What do I know?
Not much beyond two dimensions and a lot of special cases. In two dimensions the assertion is true. $K$ is a sector. $K_t$ is the translate of $K$ by a vector directed along the bisector of the angle formed by the sector's boundary. It follows that the assertion also holds for three-dimensional trihedral cones: i.e. those $K$'s in 3-space whose affine cross-section is a triangle. It seems a dissection + limit argument ought lead from here to the general 3-dimensional result but I did not get even this.

Convex geometry terminology? If the assertion is true, then the length of the translation vector taking $K$ to $K_1$ measures of the “sharpness” of the cone $K$. This length is the reciprocal of $m(K)= \max_{q \in \operatorname{int}(K)} \operatorname{dist}(q, \partial K)/\|q\|$ as $q$ varies over the interior of $K$. Does any one know the convex geometry term for this $m(K)$ or its reciprocal, the “sharpness” of $K$?

Answer 1

$K_t$ need not be a translate of $K$. Let $A=[-4,4]\times[-1,1]\subseteq\mathbb{R}^2$ and consider the convex cone $K=\{t\cdot v;t\in[0,\infty),v\in A\times\{1\}\}\subseteq\mathbb{R}^3$. Note that $\partial K$ is contained in the union of the four planes $x_2=\pm x_3$ and $x_1=\pm4x_3$, so points in $K_1$ will be points of $K$ at distance $\geq1$ of those four planes.

Clearly $K$ has a unique point with minimal $x_3$ coordinate, the point $(0,0,0)$. However, the set $K_1$ only contains points with $x_3$ coordinate at least $\sqrt{2}$ (see (*) below), and all the points $(t,0,\sqrt{2})$ for $t\in[-1,1]$ are in $K_1$ (as they are at distance $\geq1$ of the four planes mentioned above). So $K_1$ is not a translate of $K$.

(*) To see why $K_1$ only contains points with $x_3$ coordinate at least $\sqrt{2}$, note that if $X:=\{(x,y)\in\mathbb{R}^2;y>|x|\}$, then all points in $X$ at distance $\geq1$ of $\partial X$ have $y$ coordinate $\geq\sqrt{2}$; similarly, all points of $K\subseteq\mathbb{R}\times X$ at distance $>1$ of the planes $x_2=\pm x_3$ have $x_3$ coordinate at least $\sqrt{2}$.

Answer 2

Conversations with friends today led to a solution. A generic 4-sided convex polyhedral cone in 3-space led to a counterexample. To be specific, suppose the 3-dimensional convex polyhedral cone $K$ be defined by $x \ge 0, y \ge 0, z \ge 0$ and $ax + by - cz = 0$ where $(a,b,c)$ is a unit vector with $a, b , c > 0$. One verifies that the four faces of $K$ are all two-sided, each being bounded by two rays.

The polyhedron $K_t$, $t > 0$ is defined by the inequalities $x \ge t, y \ge t, z \ge t$ and $ax + by -cz \ge t$ since $x, y, z, ax + by -cz$ represent the distance from the four faces. In particular the face $z = 1$ of $K_{1}$ is coordinatized by $(x,y)$ and is characterized by the inequalities $x \ge 1, y \ge 1$ and $a x + by \ge 1 + c$. One verifies that for `most' choices of $a, b,c$ this particular face of $K_1$ is three-sided. For example, if $a = b = c = 1/\sqrt{3}$ we have $a + b - c = 1/\sqrt{3} < 1 + 1/ \sqrt{3}$ which means that the line $a x + by = 1+ c$ crosses into the interior $x > 1, y >1$ of this face of $K_1$,

We cannot translate a polyhedron all of whose faces are two-sided and end up with one having a three-sided face. So $K_1$ cannot be a translate of $K$.