4A. Two squares with side 1 are given, such that one vertex of one square is the center of the other square. What is the smallest possible value of the area of the intersecting part?
Solution. Let's consider the diagram on the right. The angles $C O D$ and $A O B$ are equal as angles with perpendicular sides, $\measuredangle O B A=\measuredangle O D C=45^{\circ}$ and $\overline{O D}=\overline{O B}$, so it follows that $\triangle A O B \cong \triangle C O D$. Therefore, the desired area is equal to the area of triangle $B O D$, i.e., it is equal to a quarter of the area of the square,

![](https://cdn.mathpix.com/cropped/2024_06_05_c86343eb2e31158b731fg-05.jpg?height=303&width=349&top_left_y=399&top_left_x=1075)
which is $\frac{1}{4}$. Since the area of the intersecting part is constant, it is also the smallest possible value.