How many equilateral hexagons of side length $\sqrt{13}$ have one vertex at $(0,0)$ and the other five at lattice points?
Source: question 32 of this contest
The vectors with integer entries with length $\sqrt{13}$ are $(\pm 2, \pm 3)$. Let $P_1 = (0,0)$ and label the rest of the points counterclockwise so that the ith point is $P_i$. It could be useful to determine exactly which point is three vertices away from $(0,0)$. But I'm not sure how to come up with such a point. Since I think the reasoning could be relatively tedious, it's fine if the answerer comes up with an explanation for just two such points that aren't "symmetrically" the same (i.e. they aren't obtained in symmetric ways).
According to the solution, the points are $(\pm 8, \pm 3), (\pm 7,\pm 2),$ etc. An explanation of how to get $(8, 3)$ and $(7,2)$ would be nice. Also, how do they know that if the third vertex is at any of $(\pm 8, \pm 3), (\pm 3, \pm 8)$, there are 7 possible hexagons? Similarly, how do they could the possibilities for the remaining cases?
Just some thoughts.
There are $24$ convex equilateral hexagons.
You need a sequence of vectors $v_0,\dots,v_5$ from $S=\{(\pm2,\pm3),(\pm3,\pm2)\}$ such that $v_0+\cdots+v_5=0$ and each of $v_0,v_0+v_1,v_0+v_1+v_2,\dots, v_0+v_1+\cdots+v_5$ are distinct. You also need $v_i\neq v_{i+1}$ to ensure that the each corner turns.
If we assume the hexagon is convex, the we can write the vectors in counter-clockwise order and get:
$$\{(3,2),(2,3),(-2,3),(-3,2),(-3,-2),(-2,-3),(2,-3),(3,-2)\}.$$ Then a convex hexagon is determined by any subset of six of these vectors which add up to $0$, in the appropriate order. But since all of these vector add to $0,$ there we have to remove one vector and it's negative.
So there are four ways to pick the six vectors.
Then we get $6$ vectors and we can take them in counter-clockwise order starting at any one of the six points.
So we get $24$ convex equilateral hexagons.
For example, if we exclude $\pm(3,2)$ and start with vector $v_0=(-2,3),$ you get:
$$(-2,3),(-5,5),(-7,2),(-5,-1),(-2,-3),(0,0).$$
If you remove the convex requirement, it will take some work.
Just the start of the non-convex case.
If $z=3+2i,$ then we can represent the vectors as:
$$v_i\in\left\{\pm z,\pm iz,\pm \frac{13}{z},\pm i\frac{13}{z}\right\}.$$
Now, a combination of these is of the form $(a+bi)z+(c+di)\overline z=0.$ But $z$ is prime in the Gaussian integers, and not a factor of $\overline z.$ so $z\mid c+di$ and thus $c^2+d^2$ must be divisible by $3.$ That means that if $(a+bi)z+ (c+di)\overline z=0$ is a combination of six vectors from this set, then $c+di=a+bi=0,$ and thus it each vector must come with its negative.
No pair can occur more than once without one $v$ being followed by another $v.$
So we must have three distinct pairs.
Any ordering of the pairs where each vector comes three steps before the negative will not be self-intersection.
So we get: $6\cdot 4\cdot 2/2=48$ of those case for each vector set. (We count each hexagon twice.)
So we are left with counting the partitions like $121323.$