Tetrahedron $ABCD$ has $AD=BC=28$, $AC=BD=44$, and $AB=CD=52$. For any point $X$ in space, suppose $f(X)=AX+BX+CX+DX$. The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.
Set $a=BC=28$, $b=CA=44$, $c=AB=52$. Let $O$ be the point which minimizes $f(X)$.
$\boxed{\textrm{Claim 1: } O \textrm{ is the gravity center } \ \tfrac {1}{4}(\vec A + \vec B + \vec C + \vec D)}$
$\textrm{Proof:}$ Let $M$ and $N$ denote the midpoints of $AB$ and $CD$. From $\triangle ABD \cong \triangle BAC$ and $\triangle CDA \cong \triangle DCB$, we have  $MC=MD$, $NA=NB$ an hence $MN$ is a perpendicular bisector of both segments $AB$ and $CD$. Then if $X$ is any point inside tetrahedron $ABCD$, its orthogonal projection onto line $MN$ will have smaller $f$-value; hence we conclude that $O$ must lie on $MN$. Similarly, $O$ must lie on the line joining the midpoints of $AC$ and $BD$. $\square$
$\boxed{\textrm{Claim 2: The gravity center } O \textrm{ coincides with the circumcenter.} \phantom{\vec A}}$
$\textrm{Proof:}$  Let $G_D$ be the centroid of triangle $ABC$; then $DO = \tfrac 34 DG_D$ (by vectors). If we define $G_A$, $G_B$, $G_C$ similarly, we get $AO = \tfrac 34 AG_A$ and so on. But from symmetry we have $AG_A = BG_B = CG_C = DG_D$, hence $AO = BO = CO = DO$. $\square$
Now we use the fact that an isosceles tetrahedron has circumradius $R = \sqrt{\tfrac18(a^2+b^2+c^2)}$. 
Here $R = \sqrt{678}$ so $f(O) = 4R = 4\sqrt{678}$. Therefore, the answer is $4 + 678 = \boxed{682}$.