# Area of a triangle whose vertices are the only lattice points in it

Let $\Delta$ be triangle in $\mathbb R^2$ such that its vertices are lattice points (points of $\mathbb Z\times \mathbb Z$) and it has no lattice points inside or in its edges. I need to show that area of $\Delta$ must be $\frac{1}{2}$.

I don't want to use this theorem.

I have done the following: Using translation and rotation and reflection I can assume that two vertices of the triangle are $(0,0)$ and $(-1,0)$ and the other vertex is of the form $(n,m)$, where $m\geq 1$.

Now clearly if the vertices of the triangle are $(0,0),(-1,0)$ and $(n,1)$ then its area is $\frac{1}{2}$. I think this is the only choice but I cannot prove this. In particular I don't know how to show a certain point is inside of a triangle or not.

Any help proving this statement is highly appreciated.

Thank you

If $m>1$, then the lines joining $(n,m)$ to the two fixed points must both cross the line $y=1$. They cannot cross at a lattice point, so they must pass in between them, and they must pass in between the same pair of lattice points; say $k-1$ and $k$ are their $x$-coordinates. That puts the point $(n,m)$ between the lines $ky-x=0$ and $ky-x=1$. Therefore, $km-n$ is between $0$ and $1$, a contradiction, since $k,m$ and $n$ are all integers.