jmc

Let $M$ and $N$ be midpoints of $\overline{AB}$ and $\overline{CD}$. The given conditions imply that $△ABD\cong △BAC$ and $△CDA\cong △DCB$, and therefore $MC=MD$ and $NA=NB$. It follows that $M$ and $N$ both lie on the common perpendicular bisector of $\overline{AB}$ and $\overline{CD}$, and thus line $MN$ is that common perpendicular bisector. Points $B$ and $C$ are symmetric to $A$ and $D$ with respect to line $MN$. If $X$ is a point in space and $X^{\prime }$ is the point symmetric to $X$ with respect to line $MN$, then $BX=A{X}^{\prime }$ and $CX=D{X}^{\prime }$, so $f(X)=AX+A{X}^{\prime }+DX+D{X}^{\prime }$. Let $Q$ be the intersection of $\overline{X{X}^{\prime }}$ and $\overline{MN}$. Then $AX+A{X}^{\prime }\ge 2AQ$, from which it follows that $f(X)\ge 2(AQ+DQ)=f(Q)$. It remains to minimize $f(Q)$ as $Q$ moves along $\overline{MN}$. Allow $D$ to rotate about $\overline{MN}$ to point $D^{\prime }$ in the plane $AMN$ on the side of $\overline{MN}$ opposite $A$. Because $\angle DNM$ is a right angle, $D^{\prime }N=DN$. It then follows that $f(Q)=2(AQ+D^{\prime }Q)\ge 2A{D}^{\prime }$, and equality occurs when $Q$ is the intersection of $\overline{A{D}^{\prime }}$ and $\overline{MN}$. Thus $\min f(Q)=2A{D}^{\prime }$. Because $\overline{MD}$ is the median of $△ADB$, the Length of Median Formula shows that $4M{D}^2=2A{D}^2+2B{D}^2-A{B}^2=2\cdot 28^2+2\cdot 44^2-52^2$ and $M{D}^2=684$. By the Pythagorean Theorem $M{N}^2=M{D}^2-N{D}^2=8$. Because $\angle AMN$ and $\angle D^{\prime }NM$ are right angles,$$(A{D}^{\prime }{)}^2=(AM+D^{\prime }N{)}^2+M{N}^2=(2AM{)}^2+M{N}^2=52^2+8=4\cdot 678.$$It follows that $\min f(Q)=2A{D}^{\prime }=4\sqrt{678}$. The requested sum is $4+678=\boxed{682}$.