8.1. On the coordinate plane, a triangle $O A B$ is drawn, where the point of intersection of the medians is at $\left(\frac{19}{3}, \frac{11}{3}\right)$, and points $A$ and $B$ have natural coordinates. Find the number of such triangles. (Here, $O$ denotes the origin - the point $(0,0)$; two triangles with the same set of vertices are considered the same, i.e., $O A B$ and $O B A$ are considered the same triangle.)
Answer: 90.

Let $M$ be the midpoint of $AB$. Then, by the property of the median, $OM = \frac{3}{2} OG$, where $G$ is the centroid. Therefore, $M$ has coordinates $\left(\frac{19}{2}, \frac{11}{2}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_dd1d5c49961b7f585dbeg-10.jpg?height=660&width=942&top_left_y=561&top_left_x=503)

Let $\left(x_{a}, y_{a}\right)$ and $\left(x_{b}, y_{b}\right)$ be the coordinates of points $A$ and $B$. Since $M$ is the midpoint of segment $AB$, we have $x_{a} + x_{b} = 19$ and $y_{a} + y_{b} = 11$. Then, for $x_{a}$, there are 18 possible natural values: $1, 2, \ldots, 18$, each of which uniquely corresponds to a natural $x_{b} = 19 - x_{a}$. Similarly, for $y_{a}$, there are 10 possibilities. In total, there are $18 \cdot 10 = 180$ possibilities for choosing point $A$, and for each of them, a suitable point $B$ is uniquely determined. According to our agreement (triangles $OAB$ and $OBA$ are considered the same), we have $180 / 2 = 90$ variants.

Note that in all the specified variants, point $A$ (and $B$) does not lie on the line $OM$, i.e., triangle $OAB$ does not degenerate (otherwise, $y_{a} / x_{a} = 11 / 19$ would hold). Thus, all the found variants are indeed valid.