Given a parallelogram $A B C D$ and a line $e$. Determine the rectangle $A_{1} B_{1} C_{1} D_{1}$ such that the lines $A A_{1}$, $B B_{1}$, $C C_{1}$, $D D_{1}$ are parallel to $e$ and the area is minimized.
Draw lines through the vertices $A B C D$ of the given parallelogram and its center $K$ parallel to $e$, and denote these lines by $a$, $b$, $c$, $d$, and $k$ respectively. The pairs $a-c$ and $b-d$ are symmetric with respect to $k$, so the two pairs of lines are either identical or one pair of lines is enclosed by the other. In the first case - as can be easily seen - the area of the desired rectangle $T=A_{1} B_{1} C_{1} D_{1}$ can be arbitrarily small, so the problem has no solution. In the second case, let us assume that the lines $a, c$ enclose the lines $b, d$ (if this were not the case, we could obtain the procedure by swapping the roles of the pairs $a-c$ and $b-d$). Furthermore, let us assume that among the lines $b, d$, $b$ is closer to $a$ (or if this were not the case, we would swap the roles of $b$ and $d$ in the following).

![](https://cdn.mathpix.com/cropped/2024_05_02_953fcb9c9da3dcb3bd03g-1.jpg?height=431&width=718&top_left_y=522&top_left_x=693)

Let $T=A_{1} B_{1} C_{1} D_{1}$ be any rectangle that meets the requirements. Since $b$ is between the lines $a$ and $d$, the segment $A_{1} D_{1}$ intersects $b$, denote this point by $B_{2}$, and the intersection of $A_{1} D_{1}$ and $c$ by $C_{2}$. If we remove the triangle $A_{1} B_{1} B_{2}$ from $T$ and attach the congruent triangle $C_{1} C_{2} D_{1}$, the area remains unchanged, so the area of $T$ is

$$
t=m \cdot B_{1} B_{2}
$$

where $m$ is the distance between the lines $b$ and $c$. Since the value of $m$ does not depend on the choice of $T$, we need to minimize the length of the segment $B_{1} B_{2}$. Denote the reflection of $A_{1}$ over the midpoint of $B_{1} B_{2}$ by $A_{2}$, then

$$
B_{1} B_{2}=A_{1} A_{2}
$$

and $A_{2}$ lies on the reflection of $a$ over $b$, denoted by $a_{2}$. $A_{1} A_{2}$ is minimized when $A_{2}$ is the projection of $A_{1}$ onto $a_{2}$, in which case $A_{1} B_{1} A_{2} B_{2}$ is a square, and the sides of $T$ form a $45^{\circ}$ angle with $e$.