# Regular polygon with vertices on grid

Constructing regular polygons on a sheet of an orthogonal lined paper - a regular maths notebook - I pondered on what would it take for the polygon to have it's vertices at the intersection of the grid lines. (We could consider the paper infinitely spanning in all directions)

The only regular polygon that you can draw with vertices on the standard integer lattice is I'm afraid the square. I won't explain how to draw that!

We might as well consider polygons drawn in the Cartesian plane with vertices at rational points, and we might as well take the centre of our polygon to be the origin. Let $P$ and $Q$ be adjacent vertices with position vectors $v$ and $w$. Then $v\cdot v=w\cdot w=a$ say and $v\cdot w=a\cos(2\pi/n)=b$ say. We need $a$ and $b$ to be rational, so that $\cos(2\pi/ n)$ must be rational. But $2\cos(2\pi /n)$ is an algebraic integer, so it must be an ordinary integer. We reduce to the cases $n=3$, $4$ or $6$.

If we can draw a regular hexagon in $\mathbb{Q}^2$ we can draw an equilateral triangle, so let's focus on that. Embed our plane in three dimensional space, and consider the vector product $v\wedge w$. Then $|v\wedge w|^2=(v\cdot v)(w\cdot w)-(v\cdot w)^2=a^2(1-\cos^2\pi/3) =\frac34 a^2$. But $v\wedge w$ is a rational multiple of a unit vector perpendicular to our plane, so $|v\wedge w|^2$ is a square of a rational. Oops!

I suppose one could look at good approximations to regular polygons drawn on the integer lattice, or regular polygons in integer lattices in dimensions $3$ or more.

• Indeed, we can find a regular triangle (and hexagon, even a tetrahedron) in $\Bbb Z^3$. – Hagen von Eitzen Apr 16 '17 at 8:39
• Equilateral triangles and hexagon can be drawn in three dimensions. Start with the plane passing through $(1,0,0), (0,1,0), (0,0,1)$. – Oscar Lanzi Apr 16 '17 at 8:42
• Sadly, it's the truth. – Hemispherr Apr 16 '17 at 9:44
• One can draw a good lattice approximation to an equilateral triangle by finding a rational approximation to $\sqrt3$, say $\frac ab$, then drawing the points $(0,0), (b,a),$ and $(2b, 0)$. The convergents of $\sqrt3$ quickly produce triangles that are indistinguishable from equilateral. For example, the convergent $\frac74\approx\sqrt3$ gives a triangle with vertices $(0,0), (4,7), (8,0)$, which has one side of length 8 and two of length $\sqrt{65}\approx 8.06$, an error of less than 1%. The next convergent, $\frac{26}{15}\approx\sqrt3$, gives a triangle that is correct to 1 part in 1800. – MJD Apr 16 '17 at 21:08
• I come across this answer of yours only after asking this question of mine. Well, I am obviously one that "could look at good approximations etc", my question is whether you, writing that line, were aware of someone doing the same before me. Thank you @Lord – lesath82 May 7 '17 at 21:43