Frankie the frog is hopping on a large, packed grid of lily pads, shown below. The pads are circular and each is a distance 1 from its nearest neighbors. (More concretely: Each pad has a diameter of 1 and they are arranged in a hexagonal lattice.) Frankie starts at (0, 0), the center of the pad labeled A. Then she hops due east to pad B at (1, 0), and from there she hops to pad C at (1.5, √(3)/2).
She wants to continue hopping in a counterclockwise, spiral-like pattern. Each of her jumps is to the center of a neighboring pad, a net distance of 1. But there are two rules her spiral must follow:
- Each next pad must be in a more counterclockwise direction (relative to spiral’s origin at pad A) than the previous pad.
- Each pad must be farther from A than the previous pad.
After a number of hops spiraling around, Frankie realizes she is, once again, due east from A. What is the closest to A she could possibly be? That is, what is the minimum possible distance between the center of the pad she’s currently on from the center of pad A?
There are only six possible directions for Frankie to jump in, namely $\pm 1$ and $( \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2} ).$ If we let $\zeta = e^{i \pi/3}$ then we see that the directions are actually all $\zeta,$ $\zeta^2,$ $\zeta^3 = -1,$ $\zeta^4,$ $\zeta^5$ and $\zeta^6 = 1.$ Let's let define $\xi = \zeta^2$ and not that $1 = \zeta - \xi.$ In this way we see that the directions that Frankie can jump in are either $\pm \zeta$, $\pm \xi$ or $\pm 1 = \pm (\zeta - \xi),$ so all in terms of $\zeta, \xi$ coordinates. While we are still establishing preliminaries, let's also see that if we have some point $z=a\zeta + b\xi$ for any $a, b \in \mathbb{Z},$ that \begin{align*}|z|^2 &= (a\zeta + b\xi) \left(\frac{a}{\zeta} + \frac{b}{\xi}\right) \\ &= a^2 + b^2 + ab\left(\zeta + \frac{1}{\zeta}\right) \\ &= a^2 + b^2 + 2ab\cos \frac{\pi}{3} \\&= a^2 + b^2 + ab.\end{align*}
Since she wants to do make the tightest possible spiral, she will want to keep going in the same direction until the next possible clockwise motion, which will be dictated by her more counterclockwise and more distant from A requirements.
Let's see this in action. We have $C = 1 + \zeta = 2\zeta - \xi,$ with $$|C| = \sqrt{2^2 + 1^2 + 2\cdot (-1)} = \sqrt{4 + 1 - 2} = \sqrt{3}.$$ Since $|C + \xi| = |2\zeta| = 2 \gt |C|,$ we have $D = 2\zeta.$.
Now we will keep adding $\xi$ steps until $\xi - \zeta = -1$ gives us an acceptable step. So let's build up the cases here. We will have steps of $2\zeta + n\xi$ for each $n = 0, 1, \dots, N_1$ where we have \begin{align*}N_1 &= \min \{ k \mid |\zeta + (k+1)\xi| \gt |2\zeta + k\xi|\} \\ &= \min \{ k \mid 1 + (k+1)^2 + k+1 \gt 4 + k^2 + 2k \}\\ &= \min \{ k \mid k \gt 1 \} = 2.\end{align*} So after we are done with $3\xi$ steps we find ourselves at $G = 2\zeta + 2\xi = (0,2\sqrt{3})$ and ready to make steps in the $-1$ direction.
Again we will keep adding $-1=-\zeta + \xi$ steps until $-\zeta$ gives us an acceptable step. So we have steps $(2-n)\zeta + (2 + n)\xi$ for $n = 0, 1, \dots, N_2$ where \begin{align*}N_2 &= \min \{ k \mid | (1-k)\zeta + (2+k) \xi | \gt | (2-k) \zeta + (2+k) \xi | \} \\ &= \min \{ k \mid (1-k)^2 + (2+k)^2 + (1-k)(2+k)\gt (2-k)^2 + (2+k)^2 + (2-k)(2+k) \} \\ &= \min \{ k \mid k^2 +k + 7 \gt k^2 + 12 \} \\ &= \min \{ k \mid k \gt 5 \} = 6.\end{align*} So we will find ourselves at $M = -4\zeta + 8\xi = (-6,2\sqrt{3}).$
Noting that the number of $-1$ steps was exactly twice the number of $\xi$ steps, we continue on we see that we take 12 $(-\zeta)$ steps followed by 24 $(-\xi)$ steps, then 48 $+1 = \zeta - \xi$ steps to arrive at $32\zeta - 64 \xi,$ ready to take new $\zeta$ steps. Eventually we will arrive at the positive $x$-axis, which is due east of the center of A, whenever we have some number of $n (\zeta - \xi) = (n,0).$ Since we are starting with $-64\xi,$ Frankie will arrive at the point $64\zeta - 64\xi = (64,0),$ which is 64 units from the center of A.
