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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?

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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
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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

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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
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@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
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