19. (BUL 2) A tetrahedron $A B C D$ is given such that $A D=B C=a$; $A C=B D=b ; A B \cdot C D=c^{2}$. Let $f(P)=A P+B P+C P+D P$, where $P$ is an arbitrary point in space. Compute the least value of $f(P)$.
19. Let $M$ and $N$ be the midpoints of segments $A B$ and $C D$, respectively. The given conditions imply that $\triangle A B D \cong \triangle B A C$ and $\triangle C D A \cong \triangle D C B$; hence $M C=M D$ and $N A=N B$. It follows that $M$ and $N$ both lie on the perpendicular bisectors of $A B$ and $C D$, and consequently $M N$ is the common perpendicular bisector of $A B$ and $C D$. Points $B$ and $C$ are symmetric to $A$ and $D$ with respect to $M N$. Now if $P$ is a point in space and $P^{\prime}$ the point symmetric to $P$ with respect to $M N$, we have $B P=A P^{\prime}, C P=D P^{\prime}$, and thus $f(P)=A P+A P^{\prime}+D P+D P^{\prime}$. Let $P P^{\prime}$ intersect $M N$ in $Q$. Then $A P+A P^{\prime} \geq 2 A Q$ and $D P+D P^{\prime} \geq 2 D Q$, from which it follows that $f(P) \geq 2(A Q+D Q)=f(Q)$. It remains to minimize $f(Q)$ with $Q$ moving along the line $M N$.
Let us rotate point $D$ around $M N$ to a point $D^{\prime}$ that belongs to the plane $A M N$, on the side of $M N$ opposite to $A$. Then $f(Q)=2\left(A Q+D^{\prime} Q\right) \geq$ $A D^{\prime}$, and equality occurs when $Q$ is the intersection of $A D^{\prime}$ and $M N$. Thus min $f(Q)=A D^{\prime}$. We note that $4 M D^{2}=2 A D^{2}+2 B D^{2}-A B^{2}=$ $2 a^{2}+2 b^{2}-A B^{2}$ and $4 M N^{2}=4 M D^{2}-C D^{2}=2 a^{2}+2 b^{2}-A B^{2}-C D^{2}$. Now, $A D^{\prime 2}=\left(A M+D^{\prime} N\right)^{2}+M N^{2}$, which together with $A M+D^{\prime} N=$ $(a+b) / 2$ gives us
$$
A D^{\prime 2}=\frac{a^{2}+b^{2}+A B \cdot C D}{2}=\frac{a^{2}+b^{2}+c^{2}}{2} .
$$

We conclude that $\min f(Q)=\sqrt{\left(a^{2}+b^{2}+c^{2}\right) / 2}$.