247. Two opposite edges of a tetrahedron have lengths $b$ and $c$, while the others are equal to $a$. What is the minimum value of the sum of the distances from an arbitrary point in space to the vertices of this tetrahedron?
247. Let in tetrahedron $ABCD$, $|AB| = b$, $|CD| = c$, and the other edges are equal to $a$. If $N$ is the midpoint of $AB$ and $M$ is the midpoint of $CD$, then the line $MN$ is the axis of symmetry of the tetrahedron $ABCD$ (Fig. 49, a).

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-123.jpg?height=260&width=356&top_left_y=538&top_left_x=256)
d)

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-123.jpg?height=306&width=373&top_left_y=534&top_left_x=625)
b)

Fig. 49.

It is now not difficult to prove that the point for which the sum of the distances to the vertices of the tetrahedron reaches its minimum value must lie on the line $MN$. Indeed, take an arbitrary point $P$ and the point $P''$ symmetric to it with respect to the line $MN$. Then the sums of the distances from $P$ and $P''$ to the vertices of the tetrahedron are equal. If $K$ is the midpoint of $PP''$ ( $K$ lies on $MN$ ), then in the triangles $PAP''$, $PBP''$, $PCP''$, and $PDP''$, $AK$, $BK$, $CK$, and $DK$ are medians, and the median of a triangle is less than half the sum of the sides enclosing it.

The value of $|MN|$ is not difficult to find,

$$
|MN| = \sqrt{a^2 - \frac{b^2}{4} - \frac{c^2}{4}} = d
$$

Consider the isosceles trapezoid $LQRS$ (Fig. 49, b), where the bases $|LS|$ and $|QR|$ are equal to $b$ and $c$, and the height is equal to $d$. Let $F$ and $E$ be the midpoints of the bases $LS$ and $QR$. If $K$ is a point on $MN$ and $T$ is a point on $FE$, such that $|FT| = |NK|$, then, obviously, the sums of the distances from $K$ to the vertices $A, B, C$, and $D$ and from $T$ to the vertices $L, S, Q$, and $R$ are equal. And in the trapezoid $LQRS$ (and in any convex quadrilateral), the sum of the distances to the vertices reaches its minimum value at the intersection of the diagonals and is equal to the sum of the diagonals.

Answer: $\sqrt{4a^2 + 2bc}$.