Can a hexagonal grid embed rectangular coordinates?

Question

I'm trying to figure out if a hexagonal grid can embed rectangular coordinates in whole numbers of "Y-steps". In the image below, one "Y-step" is the spacing between red hexagon centers in the Y dimension.

For some arbitrary hexagonal grid size, how many hexagons do I need to produce some whole-valued number of "Y-steps" in the X dimension?

Another way to ask this might be:

Select four hexagons whose centers create the corners of a square. In the hexagon grid orientation shown below, how many horizontal hexagons are needed to create such a square, and then how many vertical "steps" are needed in the Y dimension? Both X and Y values need to be whole integers.

In case it helps, this site provides great info in hexagonal coordinates, but I've not figured out how to pin down a way to solve this. We are using the "pointy top" orientation.

Answer 1

Tragically, you will never get a square.

If the side length of a hexagon is $1$ unit, then the centers of two horizontally adjacent hexagons are $\sqrt 3$ units apart. However, the distance from the center of a hexagon to the center of the next hexagon directly above it is $3$ units. If we take $a$ horizontal steps and $b$ vertical steps, we will get an $a \sqrt3$ by $3b$ rectangle, and we can never have $a \sqrt3 = 3b$, because it would give us $\frac ab = \sqrt3$, but $\sqrt3$ is irrational.

The best thing we can do is find close approximations; for example, $\frac{19}{11} = 1.72727\dots$ is not far from $\sqrt3 \approx 1.73205$, so a rectangle formed by $19$ horizontal steps and $11$ vertical steps is quite close to a square.

Answer 2

There is a neat visual proof by contradiction that a perfect square is impossible. Note that the lattice of hexagon centers has six-fold rotational symmetry about every point. That means if you rotate any lattice point 60 degrees about another lattice point, you'll get a third lattice point. Assuming we had a square $ABCD$ consisting of lattice points, we could rotate each vertex around the next and end up with a smaller square $A'B'C'D'$:

By repeating this procedure enough times, we would end up with a square smaller than the distance between neighboring hexagons, which is impossible! $\square$

I learned of this proof from Joel David Hamkins' blog.