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Is it possible to draw any regular polygon in such a way that its vertices be on the double grid, made of a rational and an irrational grid, superimposed, like in the image below?

If yes, what would be the step of the irrational grid?

(It is, evidently, possible to find such a double grid for drawing squares and equilateral triangles, as can be seen in the image below.)

Double grid, rational and irrational

Example of a double grid, rational and irrational (the step of the blue grid was taken $\sqrt 3$).

Robert Werner
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2 Answers2

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What I actually ended up doing below was merely to cover the case where certain pairs of vertices had to be at grid points of one of the grids. Allowing one vertex of the pair to be on one grid and one on the other opens up some possibilities; allowing the use of intersections of lines from different grids, along with some arbitrary offset of the grids' origins, opens up others. There might still be a way to make an argument like the following stick, perhaps using a polygon of more than ten sides. So I'll leave this here, but it is unfinished and possibly unproductive.

I interpret your question to be whether it is true for every $n \geq 3$ that there exists a regular polygon of $n$ sides that has all vertices on grid points.

Note that if $P,$ $Q,$ $R,$ and $S$ are grid points of a regular rectangular grid and the lines $PQ$ and $RS$ are parallel, then the ratio of distances $PQ:RS$ is rational. (Note that this result is true for any orientation of the grid that allows each line to intersect multiple grid points, not just the case where the lines are parallel to a line of the grid.)

Given a regular decagon $ABCDEFGHIJ$, consider the lines $AF,$ $BE,$ and $CD.$ These lines are all parallel. But the distances $BE$ and $CD$ are in the ratio $$\frac{BE}{CD} = \frac{\csc\left(3\pi/10\right)}{\csc\left(\pi/10\right)} = \frac12(3 - \sqrt5),$$ which is irrational, so the points $B$ and $E$ cannot be on the same grid as the points $C$ and $D.$

The distances $AF$ and $BE$ are in the ratio $$\frac{AF}{BE} = \frac{\csc\left(5\pi/10\right)}{\csc\left(3\pi/10\right)} = \frac14(1 + \sqrt5),$$ so $A$ and $F$ cannot be on the same grid as $B$ and $E.$

Finally, the distances $AF$ and $CD$ are in the ratio $$\frac{AF}{BE} = \frac{\csc\left(5\pi/10\right)}{\csc\left(\pi/10\right)} = \frac14(\sqrt5 - 1),$$ so $A$ and $F$ cannot be on the same grid as $C$ and $D.$

In short none of the three pairs of points can be grid points of the same grid as either of the other two pairs. You need at least three grids in order to be able to place all the points of a regular decagon on regular rectangular grids.

So if the pairs of vertices must fall on a regular grid point of one grid or the other, the answer is no, there is no mere pair of two grids that can provide vertices for regular polygons of all numbers of sides.

David K
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  • I would have given you a plus but I do not have enough reputation. Just a thought, a circle is a regular polygon with an infinite number of vertices. I guess it is impossible to place all of them on a double grid like the one I posted. – Robert Werner Jan 23 '21 at 14:24
  • What about the case when, at least some, of your P, Q, R, S points are intersections between a red and a blue line (see the image I posted). I guess that, in such a situation, PQ : RS can be irrational but I do not know whether PQ can be parallel with RS. – Robert Werner Jan 23 '21 at 14:35
  • I did not notice (until your comments hinted at it) that one of your triangle's vertices was on an intersection of red and blue grid lines. This complicates the picture, although (as you say) there is a question about when two such segments can be parallel. I don't think this solution is anywhere near complete, sorry. – David K Jan 23 '21 at 16:53
  • Somebody proposed a solution to this problem: "Knowing that $f(n)=n$, $g(n)=n\sqrt3$. Find $h(n)$ that contains all elements of $f(n)$ and $g(n)$ in ascending order." (https://math.stackexchange.com/questions/4000848/knowing-that-fn-n-gn-n-sqrt3-find-hn-that-contains-all-elements-of). Basically, any point $(x,y)$ belonging to this double grid, red and blue, has its coordinates $(h(m), h(n))$. These coordinates might help in demonstrating that only a very limited set of regular polygons can be drawn on the grid. – Robert Werner Jan 29 '21 at 09:49
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I suppose with rational and irrational grid you mean that there are a rational number $q$ and an irrational number $i$ such that for every point of the poligon, one of its coordinates is an integer multiple of either $q$ or $i$.

In this case, the answer is negative. Pick for example a square which has opposite vertices $(\sqrt{2},\sqrt{2}), (\sqrt{3},\sqrt{3})$.

Then, if this square followed what you say, $\sqrt{2}$ and $\sqrt{3}$ should be integer multiples of some irrational $i$, that is, $\sqrt{2}=ni$ and $\sqrt{3}=mi$, for some irrational $i$. However, this is impossible, because then we would have $\sqrt{3}=\frac{m}{n}\sqrt{2}$ for some integers $m,n$, which is false.

Saúl RM
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  • A rectangle is not a regular polygon. Also you are assuming that the two grids have one gridpoint in common (the origin), and I'm not sure if that is a requirement. – Jaap Scherphuis Jan 22 '21 at 13:49
  • oh sorry, I misread. I have edited it so it is a square – Saúl RM Jan 22 '21 at 13:51
  • Hm I assumed the origin had to be a point in common. If that isn´t a requirement the problem is more interesting. You should indicate the definitions in the question more clearly though – Saúl RM Jan 22 '21 at 13:55
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    Either way I think you could prove that some regular poligons centered in the origin can´t have all their vertices in two grids, using that if $p$ is prime then the $p^{th}$ roots of unity are linearly independent over $\mathbb{Q}$. But even then I don´t think it would be that easy. – Saúl RM Jan 22 '21 at 14:10
  • Definitely, drawing squares and equilateral triangles is possible. There are a lot of visible squares in the image I posted and I also added, a few minutes ago, a magenta equilateral triangle. – Robert Werner Jan 22 '21 at 15:33
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    In fact, given any 3 points $P_1,P_2$ and $P_3$, it´s easy to make a grid that passes through the 3 of them: you make it so that the horizontal lines through $P_1$ and $P_2$ are in the grid and the vertical line through $P_3$ is in the grid. Thus, given any poligon of $\leq6$ points, you can make 2 grids that pass through all its vertices.

    If you allow "inclined grids", you can get any 8 points into 2 grids.

    However, as I said above, I am inclined to believe the answer to the general problem is false, but proving it will probably require at the very least knowledge about roots of unity.

     – Saúl RM Jan 23 '21 at 00:44