In a square grid with squares of dimensions $1 \mathrm{~cm} \times 1 \mathrm{~cm}$, three nodes $K, O$, and $Z$ are marked. Determine the node $A$ such that the area of the quadrilateral $K O Z A$ is $4 \mathrm{~cm}^{2}$.

(E. Semerádová)

![](https://cdn.mathpix.com/cropped/2024_04_17_a8e752f11cb2df677c5bg-3.jpg?height=457&width=596&top_left_y=1867&top_left_x=730)

Hint. First, determine the area of the triangle $K O Z$.
The area of triangle $K O Z$ is equal to the difference between the area of the surrounding rectangle and the three corner triangles, i.e., $12-2-2-3=5\left(\mathrm{~cm}^{2}\right)$. Since the area of quadrilateral $K O Z A$ is supposed to be $4 \mathrm{~cm}^{2}$, which is less than the area of triangle $K O Z$, point $A$ must lie inside this triangle. There are four such points:

![](https://cdn.mathpix.com/cropped/2024_04_17_a8e752f11cb2df677c5bg-4.jpg?height=459&width=583&top_left_y=333&top_left_x=742)

The areas of the corresponding quadrilaterals are mutually different, so no more than one of the marked points can be correct. By a similar calculation as at the beginning, we determine that the quadrilateral $K O Z A_{1}$ has an area of $4 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_a8e752f11cb2df677c5bg-4.jpg?height=454&width=580&top_left_y=1024&top_left_x=744)

Note. The areas of the other three quadrilaterals are $1.5 \mathrm{~cm}^{2}, 3 \mathrm{~cm}^{2}$, and $4.5 \mathrm{~cm}^{2}$.

The way of cutting out the quadrilateral from the given triangle is determined by the labeling of the vertices and related conventions. In addition to quadrilateral $K O Z A_{1}$, there are other quadrilaterals with an area of $4 \mathrm{~cm}^{2}$ and with the marked points as vertices: $K O A_{1} Z, K O A_{2} Z$, and $K A_{3} O Z$.