Given the rectangle $A B C D$. On the rays $A B, B C, C D$, and $D A$ starting from $A, B, C$, and $D$ respectively, we mark the points $A_{1}, B_{1}, C_{1}, D_{1}$ such that

$$
\frac{A A_{1}}{A B}=\frac{B B_{1}}{B C}=\frac{C C_{1}}{C D}=\frac{D D_{1}}{D A}=k>0
$$

For what value of $k$ will the area of the quadrilateral $A_{1} B_{1} C_{1} D_{1}$ be minimal?
If $k \geq 1$, then the quadrilateral contains the rectangle, so its area is larger than the area of the rectangle. This area will definitely be smaller if $k < 1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_747dcbe1e44c3cf816e5g-1.jpg?height=440&width=813&top_left_y=257&top_left_x=645)

Let $AB = CD = a$, $BC = AD = b$, then from (1) we have $AA_1 = ka = CC_1$ and $BB_1 = kb = DD_1$, as well as $A_1B = DC_1 = (1-k)a$ and $B_1C = AD_1 = (1-k)b$. Let's write down the area of the quadrilateral, using the above relationships, as well as the fact that two pairs of the resulting right triangles are congruent:

$$
\begin{gathered}
t = ab - ka(1-k)b - bk(1-k)a = ab - 2abk + 2abk^2 = \\
= 2ab\left(k^2 - k + \frac{1}{2}\right) = 2ab\left[\left(k - \frac{1}{2}\right)^2 + \frac{1}{4}\right]
\end{gathered}
$$

The value of the quadratic expression in the parentheses is the smallest when $k = \frac{1}{2}$, that is, when $A_1, B_1, C_1, D_1$ exactly bisect the corresponding sides of the rectangle.