Unex-Pick-ted problem of minimum area

Using the vector formula for area of a triangle, we get

$\text{Area } = \dfrac{1}{2} \Bigg\lvert \ \vec{OA} \times \vec{OB} \ \Bigg\rvert = \dfrac{1}{2} \big\lvert \ ad - bc \ \big\rvert$

Since $a,b,c,d$ are all even, $ad - bc$ must be a multiple of $4$ , so the lowest possible non-zero value for the above expression is $\dfrac{1}{2}(4) = 2$ .

This can be attained for an infinite number of pairs of points $A$ and $B$ , [e.g. $(2,4)$ and $(4,6)$ ], including all pairs $(2,2n)$ and $(2,2n+2)$ on the line $x = 2$ as well as all pairs $(2n,2)$ and $(2n+2,2)$ on the line $y = 2$ .

Relevant wiki: Pick's Theorem

Since the numbers $a, b, c, \text{and}\; d$ are even numbers, we obtain that the boundary of the triangle contains at least the following points with integer coordinates: $(0,0), (\frac{a}{2}, \frac{b}{2}), (a,b), (\frac{a+b}{2}, \frac{c+d}{2}), (c,d), \text{and} \;(\frac{c}{2}, \frac{d}{2}).$ Now, we are going to use Pick's theorem , that expresses the area of any lattice polygon by the formula $Area(P) = I(P) + \frac{1}{2} B(P) -1,$ where $I(P)$ is the number of lattice points interior to the polygon $P,$ and $B(P)$ is the number of lattice points on the boundary of the polygon. Therefore, for the given triangle $Area=I(P) + \frac{1}{2} B(P) -1\geq I(P) + \frac{1}{2} 6 -1\geq 2$

The equality is attained by considering the triangle corresponding to $O(0,0),$ $A(2, 4)$ and $B(2, 2)$ .