Is the area of a simple polygon with integer coordinates is half an integer? I was reading shoelace formula and this occurred to me.
[Math] Properties of area of simple polygon with integer coordinates
areapolygons
Related Solutions
Let $O$ is the origin. Denote "signed area" of triangle $OAB$: $~~S_{OAB}= \dfrac{1}{2}(x_Ay_B-x_By_A)$.
It can be derived from cross product of vectors $\vec{OA}, \vec{OB}$.
If way $AB$ is $\circlearrowleft$ (if polar angle of $A$ less than polar angle of $B$), then $S_{OAB}>0$ ;
if way $AB$ is $\circlearrowright$ (if polar angle of $A$ greater than polar angle of $B$), then $S_{OAB}<0$.
Now, for each edge $A_jA_{j+1}$ ($j=1,2,\ldots,n$; $A_{n+1}\equiv A_1$) of polygon $A_1A_2\ldots A_n$ we can build $2$ vectors: $\vec{OA_j}$ and $\vec{OA_{j+1}}$.
And "signed area" of polygon (which sign depends on numerating direction) $$ S_{A_1A_2...A_n} = \sum_{j=1}^n S_{OA_jA_{j+1}} = \sum_{j=1}^n \dfrac{1}{2}(x_jy_{j+1}-x_{j+1}y_j) = \dfrac{1}{2}\sum_{j=1}^n (x_jy_{j+1}-x_{j+1}y_j). $$
When positive term adds, then square will increase, when negative, then area will decrease.
We will mark "positive" area as blue, and "negative" as red.
Illustration:
Incomplete proof
Since $P$ is convex, we can uniquely truncate $P$ with right-angled triangles with hypotenuses equal to the side lengths of $P$ and the remaining rectilinears inside.
Let reducible triangle be a right-angled triangle with Pythagorean triple $(a,b,c)$ where both $a$,$b$ and $c$ are divisible by $n$.
If all triangles above truncating $P$ is reducible triangles, area of each triangle and rectilinear is divisible by $n^2$. Thus, $2S$ must be divisible by $n$.
Please figure out the case with $P$ truncated by some non-reducible triangles.
Best Answer
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.
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.
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.