[math-fun] n-dimensional geometry puzzle

Or if you prefer, sqrt((n-1)/n) for n >= 2. Tom Tom Karzes writes:

I think the radius is just:

sqrt(n*(n-1)) / n

for n >= 2.

Tom

Tom Karzes writes:

...

I don't think the inclusion of the origin matters for n >= 2. I think you just need to find the diameter of the simplex composed of the n basis vectors (then divide by two to get the radius).

Tom

Dan Asimov writes:

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Puzzle: ------- In n-dimensional space R^n, find the radius R = R(n) of the smallest sphere containing (whether inside or on the surface) the standard basis vectors

{e_k} = {(1,0,...,0), ..., (0,...,0,1)}

and the origin 0 = (0,...,0). I.e.,

R(n) = inf {r > 0 | for some c in R^n ||p - c|| <= r for p = 0 and all p = e_k}

Apologies if this has been asked already; I don't recall.

—Dan