2
$\begingroup$

Wondering what the formula is for finding integer coordinates for an arbitrary "regular" polygon. By regular I mean symmetrical polygons like pentagon, hexagon, etc.

In particular, I would like to know what integer coordinates are for a:

  • Pentagon
  • Hexagon
  • Heptagon
  • Octagon
  • Nonagon

For instance, the Wikipedia pentagon SVG has these coordinates:

<polygon points="294,3 585.246118,214.602691 474,556.983037 114,556.983037 2.753882,214.602691" fill="white" stroke="black" stroke-width="4"/>

They are complex and unintuitive floating point values. I would instead like to figure out how to find purely integer coordinates, so for a pentagon it would be something like:

<polygon points="10,10 20,20 30,20 20,30 20,10" fill="white" stroke="black" stroke-width="4"/>

That doesn't make a pentagon, but it has all integer coordinates.

In order to accomplish this, it probably requires specifying a viewport or aspect ratio. So maybe it is wxh as 1000x1200, or 901x817 or something. I don't know how to figure this out.

To summarize, wondering what the equation is for figuring out integer coordinates for a symmetrical polygon, specifically those 5 above.

$\endgroup$
2
$\begingroup$

Those don't exist, at least not in the 2-dimensional Euclidean plane. I'm 100% sure for all of the given ones except the octagon (EDIT: see at the end why the octagon is also impossible). The reason is that for 3 points $P,Q,R$ to be consecutive vertices of such a regular $n$-gon, the $\angle PQR$ must have size $180°-360°/n$.

The tangens of the inclination $\alpha$ any straight line with equation $y=mx+n$ in a right-angled $(x,y)$-coordinate system has with the $x$-axis has is given by

$$\tan \alpha = m$$

If that line goes trough 2 points with integer coordinates (say $P=(p_x, p_y)$ and $Q=(q_x,q_y)$), we get

$$\tan \alpha_{PG} = m_{PQ}=\frac{q_y-p_y}{q_x-p_x},$$

which implies that $m_{PQ}$ is a rational number.

The same goes for the line through $Q$ and $R$ and it's inclination $\alpha_{QR}$, its tangens is also a rational number.

Now, the $\angle PQR$ we considered before is now just the difference of those two inclinations:

$$\angle PQR = \alpha_{PQ}-\alpha_{QR}$$.

The addition theorem for the tangens-function now says:

$$\tan \angle PQR = \frac{\tan \alpha_{PQ} - \tan \alpha_{QR}}{1+\tan \alpha_{PQ}\tan \alpha_{QR}}$$

The main point is now that if $P,Q,R$ have interger coordinates, this expression for $\tan \angle PQR$ is a rational number, as all the terms are rational and they are combined with elementatary operations that are closed for rationals numbers.

But $\tan 180°-360°/n$ is not a rational number for $n=5,6,7,9$. It is rational for $n=8$, as $180°-360°/8 = 135°$, so this proof does not work here.

I'm not sure what the answer is for $n=8$, could be either way.

ADDITION: It's impossible for the octagon as well. If $P,Q,R$ are the consecutive vertices of the octagon (so $Q$ is the middle one of the three), then $\angle QPR = 22.5°$. But $\tan 22.5°=\sqrt{2}-1$ is irrational, so the same argument as above applies, just for a different angle.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.