Consider four parallel lines on the plane, denoted as $a, b, c$, and $d$. We know that the distance between $a$ and $b$ is 1, the distance between $b$ and $c$ is 3, and the distance between $c$ and $d$ is also 1. Consider the rectangles whose vertices each lie on one of these four lines. How do we obtain the rectangle with the smallest possible area, and what is this area?
Solution. Let the vertices of the rectangle on lines $a, b, c$, and $d$ be $A, B, C$, and $D$ respectively. Draw perpendiculars from points $B$ and $C$ to line $a$, and let their feet be $E$ and $F$ (figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_159860851566aa67208eg-1.jpg?height=567&width=613&top_left_y=297&top_left_x=753)

The angles $\angle BAE$ and $\angle ACF$ are right angles, denoted by $\alpha$.

In the right triangle $ABE$, $BE=1$, so $AB=\frac{1}{\sin \alpha}$.

In the right triangle $ACF$, $CF=4$, so $AC=\frac{4}{\cos \alpha}$.

The area of the rectangle is:

$$
T=AB \cdot AC=\frac{4}{\sin \alpha \cdot \cos \alpha}=\frac{8}{2 \cdot \sin \alpha \cdot \cos \alpha}=\frac{8}{\sin 2 \alpha}
$$

The area of the rectangle will be minimal when $\sin 2 \alpha$ is maximal. This occurs at $\alpha=45^{\circ}$ in the range $0 \leq \alpha \leq 90^{\circ}$, where $\sin 2 \alpha=\sin 90^{\circ}=1$.

Therefore, the rectangle with the smallest area is the one where side $AB$ forms a $45^{\circ}$ angle with line $a$. In this case, the area of the rectangle is $T=8$ square units.