On squared paper, we marked a rectangle consisting of 72 squares. Then, along the grid lines, we drew a closed broken line that passes exactly once through every grid point inside or on the boundary of the rectangle and does not leave the rectangle. What can be the area of the polygon bounded by the broken line? (H)
We use Pick's theorem, which states that if a lattice polygon has $h$ lattice points on its boundary and $b$ lattice points in its interior, then its area is $T=b+\frac{h}{2}-1$. (A proof of this can be found, for example, in István Reiman's textbook "Chapters in Elementary Geometry".)

For the polygons in the problem, $b=0$, because the broken line forming the boundary of the polygon passes through each lattice point exactly once, so there are no lattice points in the interior of the polygon. If the sides of the rectangle are $a$ and $c$, then the total number of lattice points inside and on the boundary of the rectangle is $(a+1) \cdot (c+1)$, and the broken line passes through each of these points, so $h=(a+1)(c+1)$. Thus, the area of the polygon is $T=\frac{(a+1)(c+1)}{2}-1$. This must be an integer because the broken line follows lattice lines, so any small square is either completely inside the polygon or not at all. Therefore, at least one of $a$ and $c$ must be odd. Since $a \cdot c=72$, there are three possible rectangle dimensions: $1 \times 72$, $3 \times 24$, and $9 \times 8$. A polygon that meets the conditions can always be inscribed in a rectangle with one odd and one even side, as shown in the figure, so all three cases are possible.

![](https://cdn.mathpix.com/cropped/2024_05_02_2629a2854f584604f91eg-2.jpg?height=111&width=1797&top_left_y=2392&top_left_x=153)

$\frac{4 \cdot 25}{2}-1=49$, or $\frac{10 \cdot 9}{2}-1=44$.

Based on the work of Gábor Lippner (Fazekas M. Főv. Gyak. Gimn., III. o.t.)