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If I have a hexagonal lattice where distance between two adjacent points (vertices) is $1$cm, and I have a circle of large radius, then approximately how many lattice points are there inside the circle in terms of $r$, the radius of the circle? And same question for a equilateral triangle lattice, which is a hexagonal lattice with a point at the centre of each hexagon? And same questions for an $n-$ sphere in $\mathbb{R}^k$.

I think an approximation in terms of $r^2$ will always be valid, as if we move the circle around, the change in the number of points only happens to the outer layer, which has negligible points compared to the amount of points not in the outer layer...

Also, for a square lattice, I think the answer will just be $\pi r^2,$ right? But for a triangular or hexagonal lattice I'm not sure...

I know this seems like a lazy question but I just want an easy way to think about this and I don't see it. (In fact, I want to use the result for a more interesting question, but I need to know the results of these questions first...)

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  • $\begingroup$ Hint: by thinking about areas, you can bound the number of hexagons. $\endgroup$ Sep 20, 2022 at 14:27
  • $\begingroup$ The same question in $\Bbb R^k$? Do you mind to explain what would be the corresponding to an hexagonal (or triangular) lattice in $\Bbb R^k$? $\endgroup$
    – jjagmath
    Sep 20, 2022 at 15:17
  • $\begingroup$ @jjagmath maybe you are right and there isn't a standard corresponding hexagonal lattice in $R^k,$ but I think there would be a corresponding equilateral triangular one. $\endgroup$ Sep 20, 2022 at 15:23
  • $\begingroup$ @AdamRubinson There isn't one corresponding to the equilateral triangular. $\endgroup$
    – jjagmath
    Sep 20, 2022 at 16:17
  • $\begingroup$ In $\mathbb{R}^2 \cong \mathbb{C}$, don't the Eisenstein integers form a triangular lattice? $\endgroup$ Sep 21, 2022 at 3:20

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First, we ask: how many triangles fit inside the circle? The area of each triangle will be $\frac{1}{2}\frac{\sqrt{3}}{2}$, so we should have $T\approx \frac{4\sqrt{3}\,\pi r^2}{3}$ triangles in the circle.

Now it is simple to observe that there should be twice as many triangles as vertices (think of taking the top vertex of each triangle), so there should be about $\frac{T}{2}\approx \frac{2\sqrt{3}\,\pi r^2}{3}$ points.

For Hexagons, there will be $\frac{T}{6}$ inside the circle. There should be twice as many vertices, (six vertices for each hexagon, triple counted), so this gives us $\frac{T}{3}\approx \frac{4\sqrt{3}\,\pi r^2}{9}$ points.

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