[ Special cases of parallelepipeds (other). ] [ Skew lines, angle between them ]

On the diagonals $A B 1$ and $B C 1$ of the faces of the parallelepiped $A B C D A 1 B 1 C 1 D 1$, points $M$ and $N$ are taken, such that the segments $M N$ and $A 1 C$ are parallel. Find the ratio of these segments.
Consider the sections of the given parallelepiped by a plane passing through the vertices $A, D 1, B 1$ and a plane passing through the vertices $B, D, C 1$. Since the intersecting lines $A B 1$ and $A D 1$ of the first plane are respectively parallel to the intersecting lines $D C 1$ and $B C 1$ of the second plane, these planes are parallel. The ends of the segment $M N$ are located in these planes. It is known that the diagonal $A 1 C$ of the parallelepiped is divided by these planes into three equal segments, and since the segments of parallel lines enclosed between parallel planes are equal, then $M N=\frac{1}{3} A 1 C$. Suppose there is another segment $M 1 N 1$ that also satisfies the condition of the problem ( $M 1$ lies on $A B 1$, and $N 1$ - on $B C 1$ ). Then $M 1 N 1 \| M N$, so the points $M, N, M 1$ and $N 1$ lie in the same plane. Therefore, the lines $A B 1$ and $B C 1$ also lie in the same plane, which is impossible, since they are skew.

## Answer

$1: 3$

Submit a comment