সেসব ল্যাটিস ত্রিভুজের ক্ষেত্রফল যাদের ভেতরে বা উপরে কোন ল্যাটিস বিন্দু নেই

This is a special case of Pick's Theorem. In this special case, we have that if the area is $$A = I + \frac {B}{2} - 1,$$we set $I=0$ to get $A=\frac{B}{2}-1$, where $B$ is the number of lattice points on the boundary placed on the triangle's perimeter.

The goal of this post is to give an intuitive explanation of the following key special case of Pick's theorem:

Let $T$ be a ("simple") lattice triangle (i.e. triangle with vertices at lattice points) with no interior or non-vertex boundary lattice points. (In other words, the $3$ vertices of $T$ are the only lattice points on or inside $T$.) Then $T$ has area $\frac12$.

(In some sense, this special case captures all the geometry behind Pick's theorem; to prove the full Pick's theorem, we can simply combinatorially break down a general polygon into such "simple" triangles.)

What is area?

The key is better understanding what area really is.

We all know some properties that area "should" satisfy, like the additive properties. But if someone asked you to dig deeper and give an intuitive definition of area (or length, even), what would you say?

One possible definition is based on the idea of counting squares inside figures; we know what the area of a unit square should be, so if the blue lines cut out unit $1\times 1$ squares, then as a first approximation, counting blue squares tells us triangle $ABC$ has area between $15$ and $41$. How can we get a better approximation? Let's approximate using smaller squares...

As we keep iterating, the approximations get better and better. This is the intuition, which as mentioned before can be formalized---but the intuition is the important part.

In summary, we can often approximate areas of figures (like triangle $ABC$) by "almost-tiling with small building blocks" (like the $\frac18\times \frac18$ square building blocks in our last diagram). The point is that these "simple" building blocks (like squares) have "easy" areas that we understand very well.

Returning to lattice triangles with no interior lattice points

Now that we understand area better, let's return to our original problem.

Let $T$ be a lattice triangle (i.e. triangle with vertices at lattice points) with no interior or non-vertex boundary lattice points. We want to prove that $T$ has area $\frac12$.

We need to use the fact that the $3$ vertices of $T$ are the only lattice points on or inside $T$. Actually, it will be more convenient to prove that the parallelogram $P := ABDC$ (drawn below) has area $1$---this is akin to how the areas of squares and rectangles are more "natural/fundamental" than the areas of triangles. (Note that the $4$ vertices of $P$ are the only lattice points on or inside $P$.)

We will use the same tiling method as before, approximating with smaller and smaller building blocks. For instance, halving everything gives

Note that $P = ABDC$ contains exactly $1$ point of the $\frac12$-lattice $(\frac12\mathbb{Z})^2$ (i.e. point with half-integer coordinates) if we only count in its interior, and $9$ if we also include the boundary of $P$. In general, in the $\frac1n$-lattice $(\frac1n\mathbb{Z})^2$, it is not hard to show that $P$ contains exactly $(n-1)^2$ points in its interior, and $(n+1)^2$ if we also include the boundary of $P$.

Intuitively, this means $P$ is "approximately composed of around $n^2$ squares in the $\frac1n$-lattice $(\frac1n\mathbb{Z})^2$", so (since these squares have area $(\frac1n)^2$), we expect $P$ to have area $$[P] \approx n^2\cdot(\frac1n)^2 = 1,$$ which is exactly what we wanted to prove. It is possible to formalize the argument if necessary, but again, the intuition is the important part.