On "small triangles" in a square lattice

Question

A "small triangle" in a square lattice is defined as one whose vertices are non-collinear lattice points, and whose boundary and interior contain no other lattice points.

I recently came across the following:

Claim: the area of any "small triangle" is 1/2 the area of the lattice's unit cell.

I'm looking for a simple proof of this claim.

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I see that the claim would hold if the following condition holds:

Condition: the shortest side of the smallest "lattice rectangle"¹ containing the small triangle has length 1.

Then the claim follows from the elementary formula $\frac{1}{2} b h$ for the area of a triangle, since the condition implies that $b = h = 1$.

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This condition, however, does not hold for all small triangles. (In fact, the triangle defined by lattice points $(0, 0), (1, 2),$ and $(2, 3)$ is small, and the smallest lattice rectangle that contains it has sides with lengths $2$ and $3$.)

It is not clear to me that the claim would hold for all small triangles that don't satisfy the condition above.

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Update

One possibility is that the condition above should have been part of the definition of a small triangle, but was omitted by mistake. After all, the context for the claim was a proof of Pick's theorem, and it seems to me that this proof would have still gone through even if the definition of small triangles included the condition above.

IOW, if we define a "tiny triangle" as any small triangle that satisfies the condition above, then one could derive a proof of Pick's theorem from the proposition that the polygon in the theorem's premise can always be covered by a set of tiny triangles. I don't have a proof of this last assertion (anything that I can think of would probably be a tedious slog through a number of cases), but it seems to me very plausible.

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^{1 By "lattice rectangle" I mean a rectangle whose corners are lattice points, and whose sides are parallel to the sides of a lattice unit cell.}

Answer 1

In the following a small triangle is a nondegenerate triangle with vertices in ${\mathbb Z}^2$ that contains no other lattice points in its interior or boundary. Here are two proofs that such a triangle has area ${1\over2}$.

${\bf 1.\ }$Let ${\bf 0}$, ${\bf p}=(p_1,p_2)$, and ${\bf q}$ be the vertices of a small triangle $\triangle$. The lattice is symmetric with respect to the midpoint of the segment $[{\bf p},{\bf q}]$. It follows that the closed parallelogram $P$ with vertices ${\bf 0}$, ${\bf p}$, ${\bf p}+{\bf q}$, and ${\bf q}$ contains no other lattice points.

In particular ${\rm gcd}(p_1,p_2)=1$, hence there are integers $j$, $k$ such that $jp_1+kp_2=1$. Consider the map $$T:\quad {\mathbb R}^2\to{\mathbb R}^2,\qquad {\bf x}\mapsto \bar{\bf x}:=T{\bf x}$$ with matrix $$[T]=\left[\matrix{j&k\cr -p_2&p_1\cr}\right]\ .$$ This map has determinant $1$ and therefore is area preserving. Moreover $T$ maps the lattice ${\mathbb Z}^2$ bijectively onto the lattice ${\mathbb Z}^2$. It follows that the parallelogram $\bar P:=T(P)$ contains no lattice points other than its vertices. One has $$T{\bf p}=(1,0),\qquad T{\bf q}=(\bar q_1,\bar q_2)\ ,$$ whereby $\bar q_2$ is an integer which we may assume $\geq1$. The line $\bar x_2=1$ intersects the parallelogram $\bar P$ in a closed segment $\sigma$ of length $1$. It follows that $\sigma$ contains at least one lattice point ${\bf z}$, which then has to be one of the upper vertices of $\bar P$. This implies $\bar q_2=1$ and allows to conclude that $$2\,{\rm area}(\triangle)={\rm area}(P)={\rm area}(\bar P)=1\ .$$

${\bf 2.\ }$ Let ${\bf 0}$, ${\bf p}$, ${\bf q}$ be the vertices of a small triangle $\triangle$. The lattice is symmetric with respect to the midpoint of the segment $[{\bf p},{\bf q}]$. It follows that the closed parallelogram $P$ with vertices ${\bf 0}$, ${\bf p}$, ${\bf p}+{\bf q}$, and ${\bf q}$ contains no other lattice points. Its area is a natural number $\geq1$.

Claim. Any two points ${\bf x}\ne{\bf y}$ in the interior of $P$ are inequivalent modulo the lattice.

Proof. Assume that ${\bf x}=\lambda_1{\bf p}+\mu_1{\bf q}$ and ${\bf y}=\lambda_2{\bf p}+\mu_2{\bf q}$ in the interior of $P$ are equivalent modulo the lattice. WLOG we may thereby assume $$0<\lambda_1\leq\lambda_2<1,\qquad 0<\mu_1<\mu_2<1\ .$$ The point ${\bf z}:={\bf y}-{\bf x}=(\lambda_2-\lambda_1){\bf p}+(\mu_2-\mu_1){\bf q}\in P$ would then be a lattice point, but not a vertex of $P$ – a contradiction.

Corollary. Any two translated parallelograms $P+{\bf j}$, $P+{\bf k}$ with ${\bf j}$, ${\bf k}\in{\mathbb Z}^2$, ${\bf j}\ne{\bf k}$, have disjoint interiors.

Given an $N\gg1$ the $N^2$ copies $P+{\bf k}$, ${\bf k}\in[N]^2$, are therefore almost disjoint. Their union is contained in a square of side length $N+c$ with $c$ independent of $N$. It follows that $$N^2\>{\rm area}(P)\leq (N+c)^2\ ,$$ or $$2\,{\rm area}(\triangle)={\rm area}(P)\leq 1+{2c\over N}+{c^2\over N^2}\ .$$ As $N$ is arbitrary this implies ${\rm area}(\triangle)={1\over2}$.