I am wondering if for a lattice polygon an internal angle can take any value? If no which ones not and why?

I guess there will be some restrictions due to the discrete nature of the grid but I am having difficulties in determining how to use this information.

Can anyone provide a theorem and a proof?

Thank you!

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    $\begingroup$ Certainly it can only take countably many values. $\endgroup$ – Qiaochu Yuan Jul 15 '11 at 19:37
  • $\begingroup$ @Qiaochu: This is an interesting comment. I haven't thought about that (and I am not sure if it will help me) but it is fascinating nonetheless. Thank you. $\endgroup$ – vonjd Jul 15 '11 at 19:40

Any triangle with vertices in lattice points has tangents of the angles expressed by rational numbers. To do this, notice that you can split the angles of the triangles into two or more angles which can be expressed in right triangles using only the definition, and this brings rational tangents for the small angles. Then, using the formula $$ \tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$ you see that the tangent of the sum is also rational. Therefore the tangents of the angles of lattice polygons are rational numbers.

[edit] Ok. I forgot that the right angles have tangent $\infty$, so if the angles of a polygon can be angles which have rational tangents and right angles.

Here is one proof of mine that we can draw any angle with rational tangents using lattice points. My proof states that any triangle with rational tangents can be embedded in the lattice points after a certain dilatation.

Let $\tan A=\frac{a}{b}=\frac{abcd}{b^2cd},\ \tan B=\frac{c}{d}=\frac{abcd}{abd^2},\ a,b,c,d \in \mathbb{Z}^*,\ (a,b)=(c,d)=1$. Consider the orthogonal coordinate system $xOy$ with origin $D$, $Ox$ being the support of the segment $[DE]$ and points $E(b^2cd+abd^2,0),\ F(b^2cd,abcd)$. From the construction $DEF$ is similar to $ABC$ because $\tan A=\tan D,\ \tan B=\tan E$.

  • $\begingroup$ I subscribe to MartianInvader's response. $\endgroup$ – Beni Bogosel Jul 15 '11 at 20:27
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    $\begingroup$ @vonjd: It sounds like you described a right triangle whose angles have tangents 2 and $\frac{1}{2}$. Maybe you're confusing the 2-to-1 ratio of the legs with a 2-to-1 hypotenuse-to-leg ratio? The latter can't happen on the integer lattice. $\endgroup$ – MartianInvader Jul 15 '11 at 20:28
  • $\begingroup$ @vonjd: Showing that there is no lattice equilateral triangle is a golden oldie problem, with a number of solutions, including the one in the answer by Beni Bogosel. $\endgroup$ – André Nicolas Jul 15 '11 at 20:33
  • $\begingroup$ @Beni Bogosel: Surely the proof you can draw any angle wih rational tangent using lattice points is overkill. If the angle is $<90^\circ$ and has tangent $a/b$, where $a$ and $b$ are integers, start at the origin, go right distance $b$, then up $a$. For an angle $>90^\circ$ and $<180^\circ$, consider $\theta-90^\circ$. And I guess "$\infty$" is a courtesy rational. $\endgroup$ – André Nicolas Jul 15 '11 at 23:04
  • $\begingroup$ @AndréNicolas: The idea is more general than "draw an angle with rational tangents". It is more like: "embed a triangle for which all angles are rational into the integer lattice". $\endgroup$ – Beni Bogosel Dec 7 '17 at 23:36

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