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Is it possible to have a grid that contains both squares and equilateral triangles?

By grid I mean any set of the form $M \mathbb Z^2$, with $M \in GL_2\mathbb R$.


I think this is impossible, because it would contain a subgrid which is invariant under rotations of $\pi / 2$ and a subgrid which is invariant under rotations of $\pi / 3$. And I know there's no grid which is invariant under both rotations. But I can't get a conclusion from this point.

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  • $\begingroup$ How about half the plane is covered in eq triangles and the other half in squares? Is that not allowed? (At a vertex with both squares it is required that 90s + 60t = 360. Or 3s + 2t = 12. s must be even and between 0 and 4 exclusive so s = 2 and t =3. Futzing, I figured a straight vertical line separating the 2 squares from the 3 triangles was the simplest design. The only other option is two squares with a triangle in between on one side and two in between on the other. Not sure that can be extended indefiniately.) $\endgroup$ – fleablood Jul 12 '16 at 17:00
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We may view $M\Bbb Z^2$ as a subset of $\Bbb C$ and by applying a rotation and/or scaling, we may assume wlog. that $M(1,0)^T=(1,0)^T=1$. Then $M\Bbb Z=\Bbb Z + \alpha\Bbb Z$ where $\alpha=M(0,1)^T$.

Assume the lattice contains a square with vertices (in counter-clockwise order) $a,b,c,d$. Then $d-a=i(b-a)$, i.e., $M\Bbb Z$ contains two (non-zero) elements differing by a factor $i$. In other words, there exist $u,v,x,y\in\Bbb Z$, not all zero, such that $x+y\alpha=iu+iv\alpha$. Then $(x-iu)(y+iv)=(y^2+v^2)\alpha$. As the left hand side cannot be zero, we conclude $\alpha\in\Bbb Q[i]$.

Assume the lattice contains a triangle with vertices (in counter-clockwise order) $a,b,c$. Then $c-a=\omega(b-a)$ where $\omega=\frac {1+i\sqrt 3}2$, i.e., $M\Bbb Z$ contains two (non-zero) elements differing by a factor $\omega$. In other words, there exist $u,v,x,y\in\Bbb Z$, not all zero, such that $x+y\alpha=\omega u+\omega v\alpha$. Then $(x-\omega u)(y+(1-\omega)v)=(y^2+yv+v^2)\alpha$. As the left hand side cannot be zero, we conclude $\alpha\in\Bbb Q[\omega]$.

The field $\Bbb Q[i]\cap\Bbb Q[\omega]$ is a proper subfield of $\Bbb Q[i]$, hence must equal $\Bbb Q$. We conclude $\alpha\in \Bbb Q$, which would allow neither squares nor triangles in $\Bbb Z+\alpha\Bbb Z$.

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  • $\begingroup$ Thank you! Nice and clear answer. $\endgroup$ – Human Jul 13 '16 at 7:19
  • $\begingroup$ We can do the same thing with $M \mathbb{Q}^2$, can't we? $\endgroup$ – Human Jul 13 '16 at 10:01
  • $\begingroup$ @Human Yes, we can. If there were a square and a triangle in $M\Bbb Q^2$, we could multiply with the common denominator and end ed up with a square and triangle in $M\Bbb Z^2$ $\endgroup$ – Hagen von Eitzen Jul 16 '16 at 8:34
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We can cover an infinite long/finitely wide band with all triangles and another with all squares and we can cover the plane with these bands. Not sure there is any other way.

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  • $\begingroup$ Is that a grid? I mean, is that of the form $M \mathbb Z ^2$? $\endgroup$ – Human Jul 12 '16 at 17:06
  • $\begingroup$ Maybe it's not. I'm not familiar enough with the terminology. $\endgroup$ – fleablood Jul 12 '16 at 17:47

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