Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Why should such an $\alpha$ exist? – Aryabhata Jan 23 '12 at 19:22
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. – Srivatsan Jan 23 '12 at 19:31
up vote 4 down vote accepted

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

share|cite|improve this answer
Thank your for your interesting answer. Maybe it possible to find such an $\alpha$ if we don't restrict all our angles to be exact multiples of $\alpha$ but we just demand that every angle is $\epsilon$ close to a multiple of $\alpha$ for some fixed $\epsilon$? – stefan Jan 23 '12 at 19:50
@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. – André Nicolas Jan 23 '12 at 20:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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