Properties of area of simple polygon with integer coordinates

Question

Is the area of a simple polygon with integer coordinates is half an integer? I was reading shoelace formula and this occurred to me.

Answer 1

Might be the long way around this. But as a corollary to Pick's theorem which states that

$$A=i+\frac{b}{2}-1$$

where $i$ it the number of integer coordinates inside the polygon and $b$ is the number of integer coordinates on the boundary of the polygon. Because $i$ and $b$ are integers, you have $A$ in the integers or half-integers.

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And as you suggested as a result of the Shoelace formula which states that

$$A= {1 \over 2}|x_1y_2 + x_2y_3 + \cdots + x_{n-1}y_n + x_ny_1 - x_2y_1 - x_3y_2 - \cdots - x_ny_{n-1} - x_1y_n|$$

If $x_i,y_i\in\mathbb{Z}$ then the absolute value is an integer (because the sum, difference and product of integers is still an integer). And the $1/2$ our front leaves us with the integers or half-integers.

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A third approach would be to prove that any lattice triangle (a triangle with only integer coordinates) has an integer or half-integer area (hint: draw a bounding rectangle). Then show that any lattice polygon can be partitioned into non-overlapping lattice triangles.