3
$\begingroup$

Given an $n \times n$ integer grid I chose any two grid points $a,b$, draw a line $l$ through $a$ and $b$ and measure the angle between $l$ and a horizontal line. I can do this for any grid point pair and I'll get a set $A$ of angles.

I am interested in finding the largest angle $\alpha$ s.t. every element in $A$ is an integer multiple of $\alpha$.

My question is if it is possible to give a good lower bound on $\alpha$? Maybe this goes into the topic of approximating non-integer numbers using Lattices?

  • 0
    Adding to Aryabhata's comment: Such an $\alpha$ exists if and only if all the angles involved are rational multiples of a common angle, and hence iff all the angles are rational multiples of each other.2012-01-23

1 Answers 1

4

Draw a vertical line. The angle is then $\pi/2$ radians, or equivalently $90$ degrees.

Now draw the line through $(0,0)$ and $(2,1)$. The angle is now (in radians) $\arctan(1/2)$.

But it is known that $\arctan(1/2)$ and $\pi/2$ are incommensurable, meaning that there is no $\alpha$ such that each is an integer multiple of $\alpha$. So (except when our grid is very tiny) there cannot be an $\alpha$ that satisfies your conditions.

Let $\theta$ be (in degrees) a rational angle, or equivalently (in radians) a rational multiple of $\pi$. Then if $\tan\theta$ exists and is rational, we must have $\tan\theta=0$ or $\tan\theta=\pm 1$. This result goes back to Lambert, and was the main component of his proof that $\pi$ is irrational

  • 1
    @stefan: Certainly it can be done, even I can do it, crudely. Let $\alpha=10^{-6}$ (radians). Then all our angles are within $10^{-6}$ of an integer multiple of $\alpha$. The question is whether we can do much better, where the $\epsilon$ is much smaller than $1/n$. That is a possibly difficult problem in simultaneous diophantine approximations.2012-01-23