in dimensions n=4k+3, (and n=1) the answer is the regular simplex arising from n rows of mXm Hadamard matrix, where m=n+1. V(n)=m^(m/2) / n! where m=n+1. Assuming usual conjecture said matrices keep existing. In other (mod 4) dimensions, the problem will be more difficult. If you go down up to 3 dimensions to get 3 mod 4, use Hadamard to inscribe regular simplex in cube-face (of the reduced dimension) then build back up, then you will not lose out too horribly (although this will not be optimal). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)