# Pushing a convex cone and equidistants

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 $$K$$?

• TeX note: the usual TeX command for sums is \sum, not \Sigma (compare $\Sigma x_i = 0$ \Sigma x_i = 0 to $\sum x_i = 0$ \sum x_i = 0). MathJax note: backticks are parsed by the MathJax parser, not by TeX, so having too many of them in a paragraph can result in the intervening text being effectively wrapped in <pre>, as happened in this post. It's better to use " " or “ ” on MO. I edited accordingly for both. Commented May 15 at 20:34

$$K_t$$ 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 $$\sqrt{2}$$.

• Thanks! But the minimal value of $x_3$ on $K_1$ is not your $\sqrt{2}$, but rather is $x_* = \sqrt{5}/2 = 1.18..$ which is less than $\sqrt{2}$. To see this, form the sector obtained by intersecting the plane $x_2 = 0$ with your $K$. You get a sector whose angle $\theta$ satisfies $tan(\theta/2) = 2/1$ and so $sin(\theta/2) = 1/x_*$. The point $P_* = (0,0, x_*)$ is achieved by propagating the bounding rays of this sector in one unit. (Your $\sqrt{2}$ comes from the angle made by the orthogonal sector made by intersecting with the plane $x_1 = 0$.) I believe $K_1 = P_* + K$. Commented May 15 at 19:31
• It seems the point $(0,0,x_*)$ is at distance $<1$ of the point $\left(0,\frac{x_*}{2},\frac{x_*}{2}\right)$, which is in the boundary of $K$. So $(0,0,x_*)$ cannot be in $K_1$, right? Commented May 15 at 20:05
• I also made an arithmetic error. Set $x_* = \sqrt{17}/4$. The triangle with half angle $\theta/2$ has, from your measurement specifications $\tan(\theta/2) = 4/1$ whereas I had set the tangent to $2/1$ by mistake. The correct triangle is thus a $1:4: \sqrt{17}$ right triangle and $x_*$ comes from this. Commented May 15 at 22:14
• Verification: The eqns for $K_t$ can be obtained from the unit inward pointing normals of $K$. These yield the following eqns for $K_t$: $$a(y+z) \ge t$$ $$a(-y+ z) \ge t$$ $$b (x + 4z) \ge t$$ $$b (-x + 4z) \ge t$$ with $a = 1/\sqrt{2}$ and $b = 1/\sqrt{17}$. Setting $x = y = 0$ and $t =1$ and the $\ge$' to $=$' we see that $(0,0,\sqrt{17}/4)$ is on the boundary of $K_1$. Commented May 15 at 22:15
• Yes, it seemed strange to me that something like $\arctan(1/4)$ didn't appear somewhere in the argument. I didn't comment on that because the same idea still applies. E.g. the point $(0,0,\sqrt{17}/4)$ is at distance $<1$ of the point $(0,\sqrt{17}/8,\sqrt{17}/8)$, which is in the boundary of $K$ (In particular, I don't think it is in the boundary of $K_1$) Commented May 15 at 22:17

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 $$K$$.

• Does this imply that you think the solution I posted yesterday is not right? If there is some misunderstanding/you do not agree with my last comment that the point $(0,0,\sqrt{17}/4)$ is at distance $<1$ of the boundary (and so is $(0,0,x)$ for all $x<\sqrt{2}$, which is at distance $<1$ of the point $(0,x/2,x/2)$ in the boundary), we can discuss it further Commented May 16 at 11:13
• You are right Saul. I was wrong. The distance from the origin to K_1 is root 2. as you say. Your sharper angle (90 degrees) made by the plane x=0 yields the furthest point. Somehow I got the distance contributions from this angle and the other (y=0; shallower (arctan(1/4)) confused. Commented May 17 at 13:01