A rectangular parallelepiped has three edges of different lengths meeting at one vertex. We select a pair of opposite edges, then on each of the 12 edges, we find the point for which the sum of the distances to the selected edges is minimal. What is the volume of the convex body whose vertices are the points thus found?
Let the base and top faces of the cuboid $T$ be $ABCD$ and $A_1B_1C_1D_1$, respectively, and the selected edges be $AB$ and $A_1D_1$. The selection of points on the edges parallel to the selected ones (including the selected edges themselves) that minimize the sum of distances is straightforward, as the distance along these segments is constant. Thus, the desired points on the segments

$AB, A_1B_1, D_1C_1, DC$, as well as $A_1D_1, AD, BC, B_1C_1$ are clearly the endpoints of the first.

![](https://cdn.mathpix.com/cropped/2024_05_02_f048bf8680afb63e349fg-1.jpg?height=664&width=1238&top_left_y=343&top_left_x=430)

With this, we have already marked 6 vertices of $T$, and these are precisely the endpoints of 3 of the third-direction edges. Thus, if we are only concerned with determining the desired solid $K$, there is no need to look for any internal points on the edges $AA_1, BB_1$, and $DD_1$, as any such points would, due to convexity, already belong to $K$ and would not be true vertices. We also see that for any point $C^*$ on the yet unconsidered segment $CC_1$, the minimal sum of distances is achieved, and $K$ will always be formed from $T$ by cutting off the points $C$ and $C_1$, specifically the pyramids $CC^*BD$ and $C_1C^*B_1D_1$. The "horizontal" bases of these pyramids are congruent, and the sum of their heights is $CC_1$. The sum of their volumes is $CB \cdot CD \cdot CC_1 / 6$, which is $1/6$ of the volume of $T$. Thus, the volume of $K$ is $5/6$ of the volume of $T$.

For completeness, we will also determine $C^*$. It is clear that we need to measure the distances from $B$ and $D_1$ on the marked edges. We transform our problem into a planar one by folding the face $BCC_1B_1$ along $CC_1$ into the extension of the face $D_1C_1CD$. The new position of $B$, denoted as $B_2$, is connected to $D_1$ by a segment that intersects $C^*$, because for any other point $C'$ on the segment, the triangle inequality gives $B_2C' + C'D_1 > B_2D_1 = B_2C^* + C^*D_1$.

Remark. We did not need to use the fact that the edges of $T$ are of three different lengths; at most, this excluded the possibility that, for example, if $CB = CD$, we could immediately state that $C^*$ bisects $CC_1$, and that this would not be possible if we selected a different pair of edges.