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I am studying triangles on the plane $\mathbb R^2$ whose vertices have integer coordinates. If any such solid triangle (i.e., the convex hull of the vertices) has no other points with integer coordinates, we call it a primitive triangle.

I want to prove that every primitive triangle has area $1/2$. We may assume without loss of generality that one of the vertices is the origin, and the other two vertices are in the first quadrant $\mathbb N^2$. Let $(a,b)$ and $(x,y)$ be the other two vertices. Clearly, the triangle has area

$$A = \frac 12 \Vert (a,b) \times (x,y) \Vert = \frac 12 |ay - bx|$$

Thus, it suffices to show that $x,y$ are coprime and $a,b$ are coefficients arising from Bézout's identity

$$ax - by = \gcd(x,y) = 1$$

However, I have no idea how to continue from this point. Could someone give me a hint?

JoséS09
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1 Answers1

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Here is a nice algebraic proof. The points with integer coordinates form the additive group $\mathbb Z^2$. A triangle with integer coordinates can be seen as two vectors at the origin. The triangle contains no no other points with integer coordinates iff the vectors generate $\mathbb Z^2$. And this happens iff the matrix with their coordinates has determinant $\pm 1$. So, the area of the triangle is $1/2$.

lhf
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  • Forgive me for my ignorance, but is the direction $$\text{[The triangle contains no other points with integer coordinates]}\implies\text{[the vectors generate $\mathbb{Z}^2$]}$$ obvious? – Sangchul Lee Aug 26 '20 at 01:33
  • @SangchulLee: Let $v, w$ be the vectors lhf mentioned. The subgroup $G \subset \mathbb R^2$ generated by $v, w$ can be thought of the result of translating the fundamental rhombus ${ 0, v, w, v+w }$. Then $G = \mathbb Z^2$ if and only if this fundamental rhombus has no integer points besides its vertices. (cont'd) – isekaijin Aug 26 '20 at 03:19
  • @SangchulLee: Now split this fundamental rhombus into two triangles, one of which is $T_1 = { 0, v, w }$. The other triangle is itself a translated copy of $T_2 = { 0, -v, -w }$, which is a reflected copy of $T_1$, so it is clear that $T_2$ has forbidden integer points if and only if so does $T_1$. Does that work? – isekaijin Aug 26 '20 at 03:22
  • @pyon, I understand the equivalence between the triangle formulation and the parallelogram formulation. What puzzles me is the very equivalence mentioned in your first comment. Obviously we have to make a heavy use of the fact that the vertices of the parallelogram are integer points, but I am struggling to see how such a parallelogram with no interior integer points spans the entire lattice $\mathbb{Z}^2$. – Sangchul Lee Aug 26 '20 at 03:30
  • Given that all the other users seem content with this argument, I suspect that I must be missing a very obvious and straightforward idea, though. :( – Sangchul Lee Aug 26 '20 at 03:36
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    @SangchulLee: Think of $\mathbb R^2$ as the floor, and the copies of the fundamental rhombus as “tiles”. Then $G = \mathbb Z^2$ if and only if every point of $\mathbb Z^2$ is the vertex of a tile, if and only if none of the tiles has integer points besides its vertices. However, for symmetry reasons, it suffices to analyze just one fundamental tile. – isekaijin Aug 26 '20 at 03:36
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    @pyon, I see, tiling argument quite makes sense to me. Indeed, if the lattice spanned by the parallelogram misses some point of $\mathbb{Z}^2$, then the point must lie inside of one of the fundamental tiles. Thank you! – Sangchul Lee Aug 26 '20 at 03:39
  • Why do the vectors generate $\mathbb{Z}^2$ iff the matrix with their coordinates have determinant $\pm 1$? – Noppawee Apichonpongpan Mar 27 '22 at 04:45
  • @NoppaweeApichonpongpan, please ask a separate question – lhf Mar 27 '22 at 09:45