Say one wanted to test their students on the dot product formula without a calculator. One would (being a nice teacher and all) natural like to pick numbers in the plane that are "nice" and satisfy a unit circle angle, i.e. find $i,j,k,l\in\mathbb{Z}$, such that $$<i,j><k,l>=|<i,j>||<k,l>|\cdot \cos(\theta),$$ where $\theta=30,45,60,120,135,150$. After writing a script to check almost all integers within about a 15 by 15 grid, it seems as though solutions only exist for $\theta=45,135$. Has anyone see this before and/or is there a general proof of this interesting fact?
For completeness sake, here are some solutions for $\theta=45$:
<1,1>, <1,-1>
<7,1>, <4,-3>
<14,7>, <1,3>
<14,7>, <3,-1>
For angles, $30,60$, I guess I will have to resort to radical coordinates? :-(
Consider a right triangle with theta = 30, and the hypotenuse of length one. It should be easy to see the leg lengths using trig are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}.$ Thus we have an irrational value for tangent. The irrationals absorb elements of the rationals, so you will have difficulty making these work out to be integers.
On a slightly related note, are you familiar with Mathematics Educators SE? Though this question specifically probably does not fit as well there as it does here, something like "what is a good way to teach dot product without a calculator?" would probably get you god reception.
Added info: Suppose we have two vectors in $\mathbb R^2$, v=(a,b) and w=(c,d). Then, v*w=|v|*|w|*$\cos{\theta}$ where $\theta$ is the angle between them. Without loss of generality, we can think of $\theta$ as the angle of the triangle with points (0,0),(c-a,0),(c-a,d-b). Now that the above is stated for clarity, we want to say that $\theta$ is 30 (degrees) and that we want all of the points in the triangle to be in $\mathbb Z^2$. We also know that the ratio between two of the sides in an irrational number.
So let's talk about this in generality. Suppose we have x,y,z where: $x \in \mathbb Q, y \in \mathbb R, \notin \mathbb Q$ and we want to know if it is possible for z to be in $\mathbb Q$ if the relation x*y=z is true. Since $x \in \mathbb Q$, $\exists a,b \in \mathbb Z$ such that $\frac{a}{b}=x$. There also cannot be any c and d st $\frac{c}{d}=y$ because y is irrational. We also know that x*y=z $\rightarrow\frac{a}{b}*y=z$ Since all of $x,y,z \in \mathbb R$ they are closed under multiplication (and $b \neq 0$ by assumption bc $x \in \mathbb Q$), we can then say that $y=\frac{z*b}{a}$.
Suppose that $z \in \mathbb Q$. Well, z,a,b are closed under addition, so $\frac{z*a}{b} \in \mathbb Q$, which is equal to y, which is not in $\mathbb Q$. That would make life awkward. Fortunately, we have the idea of set complements, so if $z$ can't be in $\mathbb Q$, then it has to be in "not Q" (where the universe is $\mathbb R$). That by definition is the irrationals.
The conclusion to draw from this is that the irrationals "absorb," elements from the rationals. Since in the dot product question you have some integers (also rational) related by multiplication by an irrational, you cannot get a rational (or integer) solution.
Hope this explains in enough detail to make it clear.