A triangle from lattice points

The area of any lattice triangle can be found using Pick's Theorem $A = I + \frac{1}{2}B - 1$ , where $I$ is the number of interior lattice points and $B$ is the number of boundary lattice points. At a minimum, the triangle can have $0$ interior points and $3$ boundary points, which makes the minimum area $A = 0 + \frac{1}{2}3 - 1 = \frac{1}{2}$ , so $a = 1$ , $b = 2$ , and $a + b = \boxed{3}$ .

If you don't want to use Pick's Theorem, you can argue as follows:

The triangle can be inscribed in a rectangle with lattice point corners whose sides are horizontal and vertical. The area of the triangle can be computed as the area of this rectangle minus the area of several right triangles with lattice point corners whose legs are horizontal and vertical (possibly minus one other rectangle with lattice point corners whose sides are horizontal and vertical). Each of the terms in this equation is either an integer or half an integer. So the area of the triangle is either an integer or half an integer. The smallest such positive integer is $1/2.$

We can realize this minimum using the triangle with vertices $(0,0), (1,0), (1,1).$