I'm a high school teacher and someone asked me this in my class, and to be honest I'm quite stumped! I haven't done any high level math in such a long time, and I'm really not sure how to approach this.

Is the solution even approachable to a highschool student?

Thanks for the help.

  • $\begingroup$ exploring the links below, the only regular polygon you can form in the plane using lattice points is a square. $\endgroup$ – Stefan Smith Dec 27 '13 at 1:22

It is not possible, and the proof can be done with some trigonometry and a basic understanding of rational numbers:

Suppose the angle between two vectors $(a,b)$ and $(c,d)$ in the plane is $\alpha - \beta$, where $\alpha$ is the larger of the two angles made by the two vectors with the $x$-axis, and $\beta$ is the smaller.


$$\tan(\alpha-\beta) = \frac{\tan \alpha - \tan \beta}{1+\tan \alpha \tan \beta} = \frac{ad-bc}{ac+bd}$$

Thus the $\tan$ of any angle made between two integer lattice vectors is a rational number.

By dividing a regular octagon into 8 isosceles triangles and looking at the base angles of these 8 triangles, we see that we need an angle of $3\pi/8$, but $\tan (3\pi/8) = 1+\sqrt{2}$ which is irrational. Thus the octagon is not possible.

Note 1:

The angle $3\pi/8$ which is used in the proof, is actually the angle between one side of the octagon, and a "diameter" of the octagon, joining two opposite vertices. Clearly, the end-point of these two lengths would have integer coordinates if such a regular (integer) octagon were possible.

Note 2:

It seems (judging by the comment below) that my explanation has not been clear enough for everyone. I am NOT talking about the angle between two adjacent sides of an octagon being $3\pi/8$ here. The angle between adjacent sides is $3\pi/4$, and the angle I am talking about is exactly half of that.


This question and the answer by André Nicolas give more information - and an alternative proof.

  • 1
    $\begingroup$ The angle between two adjacent vertices of an octagon is $2\pi/8$, not $3\pi/8$. The tangent of this angle is not irrational, it is $1$. You can take $(a,b) = (1,0)$ and $(c,d) = (1,1)$ and these are obviously lattice vectors at the desired angle. To get an answer that works you have to include the requirement that $a^2+b^2 = c^2+d^2$ since without this the requirements are satisfiable but you get an irregular octagon. $\endgroup$ – MJD Dec 27 '13 at 9:07
  • $\begingroup$ @MJD I think you have misread or misunderstood. I am not talking about the angle between two adjacent sides of an octagon at all. Divide the octagon into 8 identical isosceles triangles, and calculate the base angle - you will find it really is $3\pi/8$. The angles of any of the 8 isosceles triangles are actually $3\pi/8$, $3\pi/8$, $\pi/4$. Draw it and see!! $\endgroup$ – Old John Dec 27 '13 at 9:16
  • $\begingroup$ @MJD Also note that your comment about it being necessary to use the lengths is also not true. I know it is possible to prove it that way, but my proof is also valid. Your example of an irregular octagon which "satisfies the requirements" most definitely does not - you don't get 8 identical isosceles triangles and you don't get a base angle of $3\pi/8$ for each triangle. $\endgroup$ – Old John Dec 27 '13 at 9:31
  • $\begingroup$ @MJD If you are still not convinced of the validity of this answer, perhaps you might look at the proof of Theorem 3, which is pretty much the same proof with a diagram included. $\endgroup$ – Old John Dec 27 '13 at 10:47
  • $\begingroup$ Note 1 makes things clearer for me (+1) $\endgroup$ – robjohn Dec 27 '13 at 13:51

Consider the two adjacent sides of a regular octagon pictured below

$\hspace{5cm}$enter image description here

First, since the sides have equal length, we have $$ \begin{align} (c-b)\cdot(b-a) &=|c-b|\,|b-a|\cos(\pi/4)\\ &=\frac{|b-a|^2}{\sqrt2} \end{align} $$ Thus, we have $$ \sqrt2=\frac{(b-a)\cdot(b-a)}{(c-b)\cdot(b-a)} $$ If all the coordinates are rational, then the quantity on the right is rational. However, since $\sqrt2$ is not rational, that is impossible.


When such an octogon exists then you are able to find a lattice point $(a,b)$ in the first quadrant and a lattice point $(c,d)$ with $d<0$ such that (i) $a^2+b^2=c^2+d^2$ and (ii) the enclosed angle is $135^\circ$.

Turning $(a,b)$ by ninety degrees counterclockwise produces the point $(-b,a)$ in the second quadrant. The condition (ii) then implies that there is a real $\lambda>0$ with $$(c,d)=-\lambda\bigl((a,b)+(-b, a)\bigr)\ .\tag{1}$$ From $\lambda={-d\over a+b}$ we conclude that in fact $\lambda\in{\mathbb Q}$.

Using the condition (i) and $(1)$ we now obtain $$a^2+b^2=c^2+d^2=2\lambda^2(a^2+b^2)\ ,$$ or $2\lambda^2=1$, which is incompatible with $\lambda\in{\mathbb Q}$.


There is a reasonably simple argument that shows that this is impossible for any regular $n$-gon other than a square (although the equilateral triangle and hexagon need a bit of extra work since these can appear on a hexagonal lattice).

The argument goes like this. Suppose there are grid points forming a regular octagon. Let $A, B, C$ be three successive vertices. Then form $D = A-B+C$ (such that $ABCD$ forms a rhombus). Now $D$ is a grid point located inside the octagon and it is not its center. Repeating this construction for the other vertices hence results in another regular octagon with vertices on grid points that is strictly smaller. This leads to an ever shrinking set of such octagons, which is not possible.

In other words, there can be no smallest such octagon and therefore there can be none at all.


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