8.1. On the coordinate plane, a parallelogram $O A B C$ is drawn, with its center located at the point $\left(\frac{19}{2}, \frac{15}{2}\right)$, and points $A, B$, and $C$ have natural coordinates. Find the number of such parallelograms. (Here, $O$ denotes the origin - the point $(0,0)$; two parallelograms with the same set of vertices are considered the same, i.e., $O A B C$ and $O C B A$ are considered the same parallelogram.)
Answer: 126.

Solution. Note that $O A B C$ is a parallelogram if and only if the point $E\left(\frac{19}{2}, \frac{15}{2}\right)$ is the midpoint of segments $O B$ and $A C$, and moreover, points $O, A, B, C$ do not lie on the same line.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bdc1b7f380f62058621g-09.jpg?height=529&width=688&top_left_y=535&top_left_x=661)

Point $E$ is the midpoint of segment $O B$, so $B$ is uniquely determined as $(19,15)$. Let $\left(x_{a}, y_{a}\right),\left(x_{c}, y_{c}\right)$ be the coordinates of points $A$ and $C$. Since $E$ is the midpoint of segment $A C$, we have $x_{a}+x_{c}=19, y_{a}+y_{c}=15$. Then for $x_{a}$ there are 18 possible natural values: $1,2, \ldots, 18$, each of which uniquely corresponds to a natural $x_{c}=19-x_{a}$. Similarly, for $y_{a}$ there are 14 possibilities. In total, there are $18 \cdot 14=252$ possibilities for choosing point $A$, and for each of them, a suitable point $C$ is uniquely determined. According to our agreement ($O A B C$ and $O C B A$ are considered the same parallelogram), we have $252 / 2=126$ options.

Note that in all the specified options, point $A$ (and $C$) does not lie on the line $O B$ (otherwise, $\left.y_{a} / x_{a}=15 / 19\right)$, i.e., the parallelogram $O A B C$ does not degenerate. Thus, all the found options indeed fit.