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Square grids

Given a finite square grid, I can rotate it and for any angle expressible as a Pythagorean triple that fits in to the grid, points on the rotated grid will align to the original.

For an infinite grid, I believe any angle atan(a/b) can be obtained where a and b are integers.

This would imply that given integers (a,b), there exists a Pythagorean triple (r.a)^2 + (r.b)^2 = c^2 where c is an integer, and r is rational?

Hexagonal grid

Q) Does this also occur on a hexagonal grid, or are the only alignments of a hexagonal grid those from rotational symmetry, at multiples of π/3?

J. M. ain't a mathematician
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Benjohn
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    Likewise, there are infinitely many triples $(a,b,c)$ that satisfy $c^2=a^2+ab+b^2$ which is the law of cosines for triangle with $\pi/3$ angle. For example (3, 5, 7); (5, 16, 19); (7, 8, 13) etc – Vasily Mitch Jul 17 '19 at 09:37
  • @VasilyMitch So … given unit vectors s and t that are 60º apart, then as an example: |3.s + 5.t| = 7? and generally, |a.s + b.t| = c given that a^2 + a.b + b^2 = c^2? – Benjohn Jul 17 '19 at 14:16

1 Answers1

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Square grid

There are some additional angles that do not come from a Pythagorean triple and yet map some grid points to grid points. For example, $5^2+5^2=7^2+1^2$, hence a suitable rotation maps $(5,5)\mapsto (7,1)$ even thow $50$ is not a perfect square. (The tangent of the rotation angle in this exampleis $\frac{5\cdot 7-1\cdot 5}{5\cdot 7+1\cdot 5}=\frac 34$)

More generally, let $A=\Bbb Z[i]$. Then for any $a,b\in A$ with $|a|=|b|>0$, a multiplication by $z:=\frac ab$ (i.e., a rotation by $\arctan\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}$) will map some grid points to grid points.

Hexagonal grid

Let $A=\Bbb Z[\omega]$, where $\omega=\frac{1+i\sqrt 3}2$. Then for any $a,b\in A$ with $|a|=|b|>0$, a multiplication by $z:=\frac ab$ will map some grid points to grid points.

Now, if $a=r+s\omega$ with $r,s\in\Bbb Z$, then $|a|^2=r^2+rs +s^2$, so these expressions take the role of the pythagorean $r^2+s^2$ of the square grid case

Hagen von Eitzen
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  • The rotation mapping $(5,5)$ to $(7,1)$ also maps $(5,0)$ to $(4,-3)$, so it does come from a Pythagorean triple. If we express points as Gaussian integers, a rotation mapping a point $z$ to a point $w$ maps $\bar{z}z$ (a real integer) to $\bar{z}w$, thus the angle comes from the Pythagorean triple formed by the coordinates and length of $\bar{z}w$. Hence, there are no such rotations whose angles are not in Pythagorean triples. – Magma Jul 17 '19 at 11:23
  • Thank you for answering! Regarding the hexagonal grid, could you please expand on the notation $Z[\omega]$? I assume its a set related to the integers? I can see $omega$ has length one – I I think it's the complex number for a rotation of $π/3$? I don't really follow the second line at all, sorry – would you be able to give a little more context for the mathematically challenged?! … I'd really like to understand the identity that generates analogous triples. – Benjohn Jul 17 '19 at 21:44