Given the unit area convex quadrilateral $A B C D$. Reflect point $A$ over $B$, point $B$ over $C$, point $C$ over $D$, and finally point $D$ over $A$. What is the area of the quadrilateral determined by these reflection points?
We denote the mirror images as $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$. From the convexity of the quadrilateral $A B C D$, it follows that the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is also convex and contains the quadrilateral $A B C D$. Let's divide the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_02_8401f01c95a9bec18860g-1.jpg?height=376&width=637&top_left_y=300&top_left_x=733)

The segment $A B$ is a median in the triangle $D B D^{\prime}$, so it divides its area into two equal parts:

$$
T_{A B D}=T_{A B D^{\prime}}
$$

$D^{\prime} B$ is a median in the triangle $A A^{\prime} D^{\prime}$, so

$$
T_{A B D^{\prime}}=T_{D^{\prime} B A^{\prime}}
$$

thus

$$
T_{A A^{\prime} D^{\prime}}=2 \cdot T_{A B D}
$$

Similarly, we can show that:

$$
\begin{aligned}
& T_{B B^{\prime} A^{\prime}}=2 \cdot T_{A B C} \\
& T_{C B^{\prime} C^{\prime}}=2 \cdot T_{C D B} \\
& T_{D C^{\prime} D^{\prime}}=2 \cdot T_{C D A}
\end{aligned}
$$

Using these, we get

$$
\begin{gathered}
T_{A^{\prime} B^{\prime} C^{\prime} D^{\prime}}=T_{A A^{\prime} D^{\prime}}+T_{B B^{\prime} A^{\prime}}+T_{C B^{\prime} C^{\prime}}+T_{D C^{\prime} D^{\prime}}+T_{A B C D}= \\
=2\left(T_{A B D}+T_{A B C}+T_{C D B}+T_{C D A}\right)+T_{A B C D}= \\
=2\left(\left(T_{A B D}+T_{B C D}+\left(T_{A B C}+T_{C D A}\right)\right)+T_{A B C D}=\right. \\
\quad=2\left(T_{A B C D}+T_{A B C D}\right)+T_{A B C D}=5 \cdot T_{A B C D} .
\end{gathered}
$$

Therefore, the area of the quadrilateral defined by the mirror points is five times the area of the original quadrilateral, which is 5 units.