Consider a triangle $ABC$ in the plane. Its vertices are all lattice points (their coordinates are integers.) Further, there are no other lattice points in this triangle, either on its boundary or interior. It can be shown in a number of ways that such a triangle has area $\frac{1}{2}$. There are several questions on this site asking for proofs, but I have an idea for an argument that I haven't seen in the answers to those questions. I'm having a little trouble formalizing it, though, and I'm not entirely sure the idea can lead to a rigorous proof.
My idea is this: Every triangle whose vertices are all lattice points necessarily has area of the form $\frac{n}{2}$ where $n$ is a positive integer (this would follow directly from the expression for the area of a triangle in terms of its vertices' coordinates.)
If a triangle $ABC$ contains a lattice point $A'$ in its edge or interior, we can construct a new triangle $A'BC$ by replacing the vertex $A$ with $A'$. Visually, it seems to me that such a triangle has strictly smaller area than $ABC$ does, and its vertices are also all lattice points, so its area is of the form $\frac{m}{2}$ with $0 < m < n$. It should then follow immediately that any such $ABC$ has area greater than $\frac{1}{2}$. It is easy to see that there are some lattice-point triangles having area $1/2$, so the only lattice-point triangles having area $\frac{1}{2}$ are those that do not contain other lattice points.
Something about this proof feels handwavy to me (in particular, I haven't justified that by replacing a vertex of a triangle with a point on its edge or in its interior, you obtain a strictly smaller triangle), but I think the basic outline is promising. Moreover, I've only established that any triangle containing lattice points has area $> 1/2$, not that every triangle not containing lattice points has area $1/2$. Can someone help with this?
Best Answer
Your approach will not work. And I think it is easier to prove the general Pick's theorem for triangles first, and then apply it to the special case where the triangle has no lattice points inside or on the boundary!
To prove the theorem for a general triangle $T$, let $R$ be a smallest-area lattice rectangle that contains $T$. Clearly each side of $R$ must have some vertex of $T$, otherwise we can shift that side inwards to obtain a smaller lattice rectangle that contains $T$. Since $T$ has $3$ vertices, at least one vertex $A$ of $T$ must be on two sides of $R$. By symmetry (of the rectangle) we can assume that $A$ is at the bottom-left corner of $R$, and that the other vertices $B,C$ of $T$ are on the top and right side of $R$ respectively. The area of $T$ can be easily expressed as the area of $R$ minus the total area of the right-angled triangles formed by the sides of $R$ and $T$.
Thus the problem reduces to proving the theorem for a right-angled triangle $ABC$ with two axis-parallel sides $AB$ and $AC$. The area of $ABC$ is half the area of the rectangle $R$ with sides $AB$ and $AC$, and it is trivial to count the number of lattice points inside or on the boundary of $R$. I shall leave the rest of the details to you. =)