The set of vertices of a polygon consists of points with coordinates $(x, y)$, where $x$ and $y$ are positive integers, and $x \mid 2 y+1$ and $y \mid 2 x+1$. What is the maximum possible area of the polygon?
First, we determine the pairs \((x, y)\) for which, in addition to the conditions, \(x \leq y\) also holds. Then \(2 x+1 \leq 2 y+1\), so from the condition \(y \mid 2 x+1\) it follows that \(2 x+1=y\), or \(2 x+1=2 y\), or \(y=1\). If \(y=2 x+1\), then \(x \mid 2 y+1=4 x+3\), which means \(x \mid 3\), so \(x=1\) or \(x=3\). The equality \(2 y=2 x+1\) cannot hold because the left side is even, while the right side is odd. Finally, if \(y=1\), then from \(x \leq y\) it follows that \(x=1\). Therefore, there are exactly three points that satisfy the condition \(x \leq y\): \((1,3)\), \((3,7)\), and \((1,1)\). Similarly, we find that there are two points that satisfy the condition \(x>y\): \((3,1)\) and \((7,3)\).

Among the polygons determined by the five points, the convex pentagon shown in the diagram has the largest area. It is easily calculated that its area is 20 square units:

\[
T=T_{1}+T_{2}+T_{3}+T_{4}=2 \cdot 2+\frac{2 \cdot 4}{2}+\frac{4 \cdot 4}{2}+\frac{2 \cdot 4}{2}=20
\]

Tran Thanh Long (Fazekas M. Fớv. Gimn., 11. o.t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_19540cb2b9f10e50dbebg-1.jpg?height=461&width=463&top_left_y=667&top_left_x=820)