A fancy explanation is below, but here's an edited simpler explanation because I think the jargon makes the problem seem inaccessible. In reality this problem is super accessible and I'm sure the answer is easy, I'm just stuck!
Imagine you have a triangle drawn in the coordinate plane, and all 3 of its vertices are integer points. $(0,0), (2,0)$, and $(0,2)$ is a simple example.
Now blow up the triangle by a factor of $n$, where $n$ is a positive integer. So using the coordinates above and $n=5$, our new triangle has vertices $(0,0), (10,0),$ and $(0,10)$.
Now choose any integer point on the edges or interior of this new triangle. I want to prove that you can express this integer coordinate as a sum of $n$ integer interior or edge points of our original triangle.
So let's say we choose $(5,4)$ as a point in the interior of the triangle $(0,0)$, $(10,0)$ and $(0,10)$.
$(5,4) = (1,0)+(1,1)+(1,1)+(1,1)+(1,1)$.
So we have $(5,4)$ expressed as the sum of exactly $5$ points from our original triangle.
It is fact that we can always do this. But why?
Here's the original, fancy explanation:
A convex lattice polytope $P$ is normal if given any positive integer $n$, every integer lattice point of the dilation $nP$, obtained from $P$ by scaling its vertices by the factor $n$ and taking the convex hull of the resulting points, can be written as the sum of exactly $n$ lattice points in $P$.
Let $P$ be a polygon with integer vertices. I'd like to show that:
$n \in \mathbb N, z \in nP \cap \mathbb Z^{2} \implies \exists x_{1},..., x_{n} \in P: x_{1}+...+x_{n}=z$
I know that every polygon is normal, but I can't come up with a convincing argument as to why this is the case.
Given a polygon $P$, we can triangulate it until we are left with simplices that cover the polygon. So I believe the problem is equivalent to showing that any arbitrary simplex in $\mathbb R^{2}$ with integer vertices is normal.
But this is where I get stuck. It's clear to me that a specific simplex, say formed by $[(0,0), (0,1), (1,0)]$, can be shown to be normal. But what's the step needed to show an arbitrary simplex is normal?
Or maybe this is the wrong approach?
Thanks!