Determining the Largest Face of a Simplex This question is in the vein of my former question Fast Comparing of the Volume of Simplices Defined by Sidelengths, but it has a different twist, that may allow for an easier answer: 

Questions: 
  
  
*
  
*Given a non-degenerate simplex, that is defined via its sidelengths, how can the face with the largest "hyper area" be found efficiently and, 
  
*are there counterexamples to the conjecture, that the vertex with smallest length-sum of adjacent edges isn't contained in the largest face's containing hyperplane?  
  

edit:
calculating the area of each face and then taking the maximum of those values is of course a valid method, but I am hoping for something more efficient because the problems are of very high dimension.
This edit is motivated by the valuable feedback of @IgorRivin.
 A: I am not sure what you mean by "efficiently", but the Cayley-Menger determinant can be used to compute the volume of a face in time $O(d^3),$ and since there are $O(d)$ maximal dimension faces, the maximum can be computed in time $O(d^4).$ As for the second question, do you even know this to be true in three dimensions?
A: Surprisingly it is possible to determine the largest face of a $d$-simplex in $O((d+2)^3)$ (or more precise: the problem isn't harder than in inverting a $(d+2)\times (d+2)$-matrix, namely the associated Cayley-Menger matrix $\hat{B}$.  
I was able to find an efficient solution of the stated problem, when I discovered an online article by G. Westendorp, dated April 2013 A formula for the N-circumsphere of an N-simplex.
In that article one finds an interpretation of the inverse of a Cayley-Menger matrices, $\hat{B}$$^{-1}$, that makes them (at least to me) by orders of magnitudes more interesting, than $\hat{B}$ itself.  

I was able to find an efficient solution to the problem, when I read that $\hat{B}$$^{-1}$ contains as entries the diameter and the barycentric coordinates of the simplexe's circum-sphere and center.  
The key observation that occured to me was that the barycentric coordinate of a vertex $v$ corresponds to the hyper-volume of the simplex obtained by replacing $v$ with the circum-center. The resulting "center-simplex" contains all edges of the face, that is opposite to $v$, plus $d$ edges of length $r$, the radius of the circum-sphere.   
As the same is true for all other simplex-vertices, the relative order of the volumes of the associated "center-simplices" is solely owed to the hyper-area of the "base-face", i.e. the one that doesn't contain the circum-center of the original simplex.  

All together the answer to the problem of efficiently determining the largest face of a simplex is:
  
  
*
  
*calculate the inverse $\hat{B}$$^{-1}$ of the Cayley-Menger matrix, that is associated with the simplex
  
*find in the first row/column the smallest barycentric coordinate $\alpha$ of the circumcenter  
  
*report the face opposite to the vertex with barycentric weight $\alpha$   
  

All in all it was possible to reduce the complexity by a linear factor, whereas normally, with divide and conquer, it is only possible to replace a linear factor by a logarithmic one.
