Minimum Area of Triangle whose Vertices are Lattice Points

Theorem

Let $T$ be a triangle embedded in a cartesian plane.

Let the vertices of $T$ be lattice points which are not all on the same straight line.

Then the area of $T$ is such that:

$\map \Area T \ge \dfrac 1 2$

Proof

Without loss of generality let one of the vertices of $T$ be at $\tuple {0, 0}$.

Let the other $2$ vertices be at $\tuple {a, b}$ and $\tuple {x, y}$.

By Area of Triangle in Determinant Form with Vertex at Origin‎:

$\map \Area T = \dfrac {\size {b y - a x} } 2$

As the vertices of $T$ are non-collinear, $\map \Area T \ge 0$.

Thus $\size {b y - a x} > 0$.

As $\tuple {a, b}$ and $\tuple {x, y}$ are lattice points, all of $a, b, x, y \in \Z$.

Thus $\size {b y - a x} \ge 1$.

Hence the result.

$\blacksquare$

Sources

• 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-3}$ The Linear Diophantine Equation: Exercise $6$