Picking Coordinates 2

$N=2021 \cdot \frac{2021+1}{2}$ - number points in equilateral big triangle, consisting of $2021$ rows of unit equilateral triangles. $M=\frac{2021+1}{2}$ - number of points on the median of a big triangle.

Answer $\frac {1}{6 } (N+5M)=341381$

Let's first count the total number of vertex points. Above the top row (row 1) there is 1 vertex, above row 2 there are 2,... and finally below the bottom row (row n) there are n+1. When having n rows, the total number of vertices is $v=1+2+...+n+(n+1) = \frac{n^2+3n+2}{2}$

Of all these vertices, some are private to a right-triangular region, some are shared with other regions. Let's investigate this. Note:

• On a line dividing the triangle in two mirror images, there are $1+ \lfloor \frac{n}{2} \rfloor$ vertices
• There are three such lines and all vertices that are shared among regions are on such a line
• There is a vertex in the centre exactly if n is a multiple of 3

We can now classify the vertices:

• Shared by 6 right triangles: $v_6=1$ if $n$ is an integer multiple of 3, $v_6=0$ otherwise
• Shared by 2 right triangles: $v_2=3(1+ \lfloor \frac{n}{2} \rfloor-v_6)$
• All others are private to some right triangle: $v_1=\frac{n^2+3n+2}{2}-v_2-v_6$

Each of the 6 right triangular areas now contains or touches x vertices where $x=\frac{v_1}{6}+\frac{v_2}{3}+v_6$

Sanity check for small n and find the answer:

If you like a closed expression for x: $x=\frac{n^2+3n+8 +6 \lfloor \frac{n}{2} \rfloor + 4\lfloor \frac{n}{3} \rfloor - 4\lfloor \frac{n-1}{3} \rfloor}{12}$

For further reference, also see the OEIS