Angle between a point in a convex polytope and the nearest point of a face Let $P \subset \mathbb{R}^d$ be a convex polytope, and let $F$ be a face of $P$ (of co-dimension 1, let's say). Now let $x \in P \setminus F$ and let $y \in F$ be the nearest point of $F$ to $x$. Then the angle between the line $xy$ and $F$ is bounded away from zero (i.e., it is at least some constant depending on the directions of the faces of $P$).
This seems like it ought to be straightforward to prove, but the proof that I've written down is a page and a half long. It also seems like it ought to be something that's well-known, but I have not been able to find a reference. Does anyone have a quick proof and/or a reference?
 A: Let $H_F\subset \mathbb{R}^d$ be the hyperplane containing $F$, and let $\pi_F\colon \mathbb{R}^d\rightarrow H_F$ be the orthogonal projection.  Set $y_0=\pi_F(x)$.  Then I claim that $y$ is the point in $P\cap H_F=F$ that is closest to $y_0$.  Indeed let $y'\in F$ be the closest to $y_0$; then if $y\neq y'$ Pythagorean theorem says $d(x,y)=\sqrt{d(x,y_0)^2+d(y_0,y)^2}> \sqrt{d(x,y_0)^2+d(y_0,y')^2}=d(x,y')$.  Now there are two cases to consider:  either $\pi_F(x)\in F$, in which case $y=y_0$, in which case the line $xy$ is orthogonal to $F$ (hence the angle is non-zero), or if $\pi_F(x)\notin F$, in which case $y\neq y_0$, in which case the angle between the line $xy$ and $F$ is the angle at $y$ in the triangle formed by $x$, $y$, and $y_0$ (hence again the angle is non-zero).  It seems intuitively clear to me that the critical angle here would have to be the minimum of the angles between the normals of adjacent facets and the normal of the facet $F$, but I don't have a proof of this.
Edit:  For a proof of this lower bound, I guess the key observation is that if $y_0\notin F$, then $y$ must lie on the boundary of $F$.  Then if I imagine the polytope sitting upright on its facet $F$, and $y$ any point on the boundary of $F$, I can imagine taking a small right triangle sitting with one side on $H_F$ and the other side orthogonal to $H_F$, and then sliding its nose, with angle $\theta$, up to $y$; if the triangle does not meet $P$ in any other point, then there can be no point $x\in P$ for which $y$ is the closest point in $F$ whose line $xy$ makes that angle $\theta$ with $F$.  If I choose my triangle so that its nose angle $\theta$ is shallower than the angle that any other facet that meets $F$ makes with $F$, then I suppose that angle $\theta$ will be a lower bound for the angle in question.
A: Needed this too and could not find a proof. Here is a proof (probably a better proof exists):
I'll assume the polytope $P$ is bounded though this can likely be extended to unbounded polytopes.  Note that any bounded convex set is a convex combination of its extreme points and for a polytope there are a finite set of extreme points. Informally these are the "corners". (Most intro courses on linear programming prove this, e.g. Introduction to linear optimization, Bertsimas and Tsitsiklis.)
Without loss of generality, I'll suppose that the face $F$ is contained in the plane $c^\top x= 0$ where $c$ is a unit vector and all points in $P\backslash F$ are such a that $c^\top x >0$. A rephrasing of the above result is that we want to show that there exists $\kappa > 0$ such that for all $x \in P\backslash F$
$$
\frac{c^\top (x-y)}{ ||x-y||} \geq \kappa ,
$$
where $y$ is the projection of $x$ onto $F$.
Let $x'\in P$ be the extreme point for which $c^\top x'$ is smallest while still positive. Define $C':=c^\top x' >0$ and $P' := \{ x \in P : c^\top x \geq C' \}$.
Both $P'$ and $F$ are two bounded polytopes that are disjoint. The $D$ be the largest distance between two points in $P'$ and $F$. We now argue we can take
$$
\kappa = \frac{C' }{ D}.
$$
There are no extreme points in between $P'$ and $F$. In other words every point $x \in P$ must be convex combination of a point in $P'$ and a point in $F$. Suppose $x \in P \backslash F$, then
$$
x = (1-p) x_0 + p x_1
$$
with $x_0 \in F$ and $x_1 \in P'$ and $p>0$. Note that if $y$ is the projection of $x$ onto $F$ then
$$
\frac{c^\top x }{ || x-y|| } \geq \frac{ c^\top (x-x_0) }{ ||x-x_0|| } = \frac{ c^\top (x_1 - x_0) }{ ||x_1 - x_0|| } \geq \frac{C'}{D} = \kappa
$$
(The first inequality uses that $c^\top x_0=0$ and that the distant from $x$ to $x_0$ must be bigger than to $y$.) This gives the required result.
