# A final incident in the flea circus: Part 1

The ringmaster of a flea circus draws a square $ABCD$ with corners $A=(+1,+1)$, $B=(+1,-1)$, $C=(-1,+1)$, $D=(-1,-1)$ in the Euclidean plane and picks a point $P$ with integer coordinates outside the square. The ringmaster then places a single flea onto point $P$. Whenever the ringmaster shouts "Hop!", the flea looks around itself and picks a corner $y$ of the square:

• If there is a unique rightmost corner (from the flea's current point of view), then $y$ becomes this unique rightmost corner.
• If the rightmost corner is not unique (as the flea is sitting on the same line with two neighboring corners $y_1$ and $y_2$ of the square), then $y$ is chosen as the rightmost corner that is farther away from the flea.

The flea then jumps over the picked corner $y$ to the mirror point on the other side. (In other words, the flea sitting in point $x$ may jump over corner $y$ to the new point $z$, so that $y$ is the midpoint between $x$ and $z$.)

Question: Is it possible that after a finite number of jumps the flea reaches a point $Q$, so that the distance between origin and $Q$ is at least $30$ times the distance between origin and $P$?

Here's the section of the plane around the square, color-coded by which point the flea will jump over:

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(The $x$ axis points rightwards and the $y$ axis points upward.)

We can model the process mathematically as

$$f(x, y) = \begin{cases} (\phantom{-}2-x,\; \phantom{-}2-y) & \quad \color{red}{x \geq 1\phantom{-} ~\text{ and }~ y < 1\phantom{-}} \\ (\phantom{-}2-x,\; - 2-y) & \quad \color{orange}{x < 1\phantom{-} ~\text{ and }~ y \leq -1} \\ ( - 2-x,\; - 2-y) & \quad \color{blue}{x \leq -1 ~\text{ and }~ y > -1} \\ ( - 2-x,\; \phantom{-}2-y) & \quad \color{green}{x > -1 ~\text{ and }~ y \geq 1\phantom{-}} \end{cases}$$

where the first case is a reflection over point ${A(1, 1)}$ if $(x, y)$ is in the red part of the plane; the second case is a reflection over point ${B(1, -1)}$ if $(x, y)$ is in the yellow part of the plane, etc.

Through lots of case splitting, we can find a similar description of $f^2$. That is, we can divide the entire plane (minus the square) into eight regions, and specify which transformation will apply to the flea's coordinates after jumping twice from its current position. They are:

$\hskip2in$

$$f^2(x,y) = \begin{cases} (x-4, y-4) & \quad \color{red}{ x \geq 3 ~\text{ and }~ y < 1} \quad \text{(region 1)}\\ (x-4, y\color{white}{+0}) & \quad \color{red}{ x \in [1, 3) ~\text{ and }~ y < 1} \quad \text{(region 2)}\\ (x-4, y+4) & \quad \color{orange}{ x < 1 ~\text{ and }~ y \leq -3} \quad \text{(region 3)}\\ (x\color{white}{+0}, y+4) & \quad \color{orange}{ x < 1 ~\text{ and }~ y \in (-3, -1]} \quad \text{(region 4)}\\ (x+4, y+4) & \quad \color{blue}{ x \leq -3 ~\text{ and }~ y \geq -1} \quad \text{(region 5)}\\ (x+4, y\color{white}{+0}) & \quad \color{blue}{ x \in (-3, -1] ~\text{ and }~ y \geq -1} \quad \text{(region 6)}\\ (x+4, y-4) & \quad \color{green}{ x > -1 ~\text{ and }~ y \geq -3} \quad \text{(region 7)}\\ (x\color{white}{+0}, y-4) & \quad \color{green}{ x > -1 ~\text{ and }~ y \in (-3, -1]}\quad \text{(region 8)} \end{cases}$$

We will prove that making repeated double jumps starting from $(4k, 1)$, where $k$ is a positive integer, leads us to $(4k+4, 1)$. Specifically, this will happen after $4k+4$ double jumps (or $8k+8$ jumps in total.)

• First, $(4k, 1)$ is (right on the boundary of) region 8.
• After one region 8 double jump (south), we are in $(4k, -3)$, in region 1.
• $k$ region 1 double jumps (south-west) later, we end up in $(0, -3-4k)$, in region 3.
• $k+1$ region 3 double jumps (north-west) later, we end up in $(-4k-4, 1)$, in region 5.
• $k+1$ region 5 double jumps (north-east) later, we end up in $(0, 4k+5)$, in region 7.
• $k+1$ region 7 double jumps (south-east) later, we end up in $(4k+4, 1)$, back on the boundary of region 8.

The final path looks like this (in purple):

$\hskip2in$

In total we made $4k+4$ double jumps in a diamond-shaped path around the origin. (Under the taxicab geometry, the first step is the one that actually brings us farther from the origin, while the others don't change the taxicab distance from the origin at all.)

Thus, we can start at $(4, 1)$, make 16 jumps to end up at $(8, 1)$, make 20 more jumps to end up at $(12, 1)$, etc., and go as far as we like. Specifically, we reach $(124, 1)$ after 3960 jumps, where the ratio of distances is

$$\frac{\sqrt{124^2+1}}{\sqrt{4^2+1}} = 30.075395... > 30.$$

Yes it is possible.

We will first position our flea at position $(+3,+1)$ and after each jump we will rotate the plane around the origin to get our flea back to positive coordinates. With this our flea will always have to jump over corner (-1,+1) and although it might sound strange to do so the flea will keep its relative position to the origin and the four corners and make this solution easier to understand.

Our first jump will be to point $(-5, +1)$ rotate 90 degrees to get to position $(+1, +5)$.

Until we are back at a position $(?, +1)$ we can keep jumping and rotatin 180 degrees and with each such jump our position gets changed by the vector $(+2, -2)$.

Our first jump for example will get us to position $(-3, -3)$ which rotated 180 degrees gives us $(+3, +3)$.

Next jump to $(-5, -1)$ rotated $(+5, +1)$.

At this point we can already see a pattern. Each time we reach a position $(?, +1)$ our position will have changed by the vector $(+2, 0)$.

Since we can repeat this infinitely we will slowly get further and further away from the origin and up to infinite distance, which is obviously more than 30 times the original distance.

• 30/2=15 so 15 times. – warspyking Nov 3 '15 at 20:35
• @warspyking The original distance is actually slightly over 3 so we would need to get to a distance of almost 95 which happens at around 46 repeats. – The Dark Truth Nov 4 '15 at 12:52