TITLE: Which values can attain the minimum solid angle in a simplex
QUESTION [9 upvotes]: Given a simplex $S$ with a vertex $v$ by the solid angle at this vertex I mean the value $\hbox{vol}(B \cap S)/\hbox{vol}(B)$ where $B$ is a small enough ball centered at $v$ (for example, in the plane, 1/4 corresponds to $90^°$). Let $\rho_{min}(S)$ denotes the minimum of the solid angles of $S$. 

My question is what values can $\rho_{min}(S)$ attain when it ranges over simplices in dimension $d$?

Comments: Since it is easy to let $\rho_{min}(S)$ to be close to $0$, the question above is equivalent with asking what is the least upper bound on $\rho_{min}(S)$.
In the plane, the question is trivial (the least upper bound is 1/6), thus more interesting cases occur when $d \geq 3$. In higher dimensions, it is relatively easy to show that $\rho_{min}(S) \leq \frac1{2(d+1)}$. But I suspect that this value should be much smaller.
A natural candidate for the least upper bound is the size of the solid angle in the regular simplex. Is this value the least upper bound?
Actually, I would be also happy with a weaker upper bound (than the one from the regular simplex) such as $\rho_{min}(S) \leq \frac{1}{2^d}$ (this is the solid angle in the $d$-cube).

REPLY [6 votes]: This answer is based on a comment by J. Kynčl. It gives a bound roughly $\rho_{min}(S) \leq \left(\frac{1}{1.074}\right)^d$ which is already exponentially decreasing, but perhaps still far from the optimum. Thus, a better bound would still be of interest.
Let $AB$ one of the longest edges of $S$. Without loss of generality, at least half of the remaining vertices of $S$ are not farther from $B$ than from $A$. Let $V_1, \dots, V_k$ be such vertices ($k \geq (d-1)/2$) and $U_1, \dots, U_{\ell}$ be the remaining vertices (closer to $A$ than to $B$). As J. Kynčl points out, the angles $V_iAB$ are at most $60^°$. We also need that angles $U_iAB$ are at most $90^°$ since $BU_i$ is at most as long as $AB$.
Let $h$ be the hyperplane perpendicular to $AB$ passing through $A$ and $h^+$ be the halfspace containing $B$ with the boundary hyperplane $h$. Let $C$ be the cone with apex $A$ determined by $S$ (that is, our task is to determine which fraction of $C$ belongs to a ball $B(A,\varepsilon)$ with center $A$ and small radius $\varepsilon$. From the discussion above it follows that $C$ is fully contained in $h^+$. Furthemore, let $\kappa$ be the affine $(k+1)$-space determined by $A$, $B$, $V_1$, $\dots$, $V_k$. We also need another $(k+1)$-dimensional cone $C_{60}$ which is formed by all points $X$ in $\kappa$ such that the angle $XAB$ is at most $60^°$. From the discussion above it follows that $V_i \in C_{60}$, and consequently $C \cap \kappa \subseteq C_{60}$. It is not too difficult to compute that $\hbox{vol}_{k+1}(B(A,\varepsilon) \cap C_{60})/\hbox{vol}_{k+1}(B(A,\varepsilon)) \leq \left(\frac{\sqrt 3}2\right)^{k+1}$, where $\hbox{vol}_{k+1}$ is the $(k+1)$-dimensional volume in $\kappa$. (See the left picture below.)

Now, let us consider a small enough ball $B(A, \varepsilon)$ and let us estimate $\hbox{vol}(B(A, \varepsilon) \cap C)/ \hbox{vol}(B(A, \varepsilon))$.
For this let us consider a $(k+1)$-space $\kappa'$ parallel with $\kappa$. The task is to show that 
$$(*) \hskip{2cm} \frac{\hbox{vol}_{k+1}(B(A, \varepsilon) \cap C \cap \kappa')}{\hbox{vol}_{k+1}(B(A, \varepsilon) \cap \kappa')} \leq \left(\frac{\sqrt 3}2\right)^{k+1}.$$
As soon as we show $(*)$ we get the same bound on $\hbox{vol}(B(A, \varepsilon) \cap C)/ \hbox{vol}(B(A, \varepsilon))$ by the Fubini theorem.
In order to show $(*)$, let us first reailize that $C \cap \kappa'$ is either empty or it equals to  $(C \cap \kappa) + Y$, where $Y$ is the intersection point of $\kappa'$ and the $\ell$-dimensional cone determined by $A$ and $U_1, \dots, U_\ell$ (here, for simplicity, we assume that $A$ is the origin). Thus, in particular, $Y \in h^+$ and $C \cap \kappa' \subseteq Y + C_{60}$. (See the right picture above.)
The final step is thus to show that $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Y + C_{60})) \leq \hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Z + C_{60}))$ where $Z$ is the center of $B(A, \varepsilon) \cap \kappa'$. This inequality is best shown by the picture below. (First shift $Y$ to $Y'$ on $h$. Then bound $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Y' + C_{60}))$ by decomposing $B(A, \varepsilon) \cap (Y' + C_{60})$ into two parts as on the middle and right picture.)

Finally, $\hbox{vol}_{k+1}(B(A, \varepsilon) \cap (Z + C_{60}))/\hbox{vol}_{k+1}(B(A, \varepsilon)) \leq \left(\frac{\sqrt 3}2\right)^{k+1}$ as in the case of $C_{60}$. This gives the final bound
$$\rho_{min}(S) \leq \left(\frac{\sqrt 3}2\right)^{(d+1)/2} \leq \left(\frac{1}{1.074}\right)^d.$$