Asian Pacific Mathematics Olympiad

Let Zillis be $d$ kilometers away from Argovia, where $d$ is a positive real number. For simplicity, we will position Argovia at $(0,0)$ and Zillis at $(d, 0)$, so that the car starts out facing east. We will investigate how the car moves around in the period of travelling the first $\ell r$ kilometers, the second $\ell r$ kilometers, . . ., and so on. We call each period of travelling $\ell r$ kilometers a section. It is clear that the car will have identical behavior in every section except the direction of the car at the beginning.

Case 1: $\ell-r \equiv 2(\bmod 4)$. After the first section, the car has made $\ell-1$ right turns and $r-1$ left turns, which is a net of $2(\equiv \ell-r(\bmod 4))$ right turns. Let the displacement vector for the first section be $(x, y)$. Since the car has rotated $180^{\circ}$, the displacement vector for the second section will be $(-x,-y)$, which will take the car back to $(0,0)$ facing east again. We now have our original situation, and the car has certainly never travelled further than $\ell r$ kilometers from Argovia. So, the car cannot reach Zillis if it is further apart from Argovia.

Case 2: $\ell-r \equiv 1(\bmod 4)$. After the first section, the car has made a net of 1 right turn. Let the displacement vector for the first section again be $(x, y)$. This time the car has rotated $90^{\circ}$ clockwise. We can see that the displacements for the second, third and fourth section will be $(y,-x),(-x,-y)$ and $(-y, x)$, respectively, so after four sections the car is back at $(0,0)$ facing east. Since the car has certainly never travelled further than $2 \ell r$ kilometers from Argovia, the car cannot reach Zillis if it is further apart from Argovia.

Case 3: $\ell-r \equiv 3(\bmod 4)$. An argument similar to that in Case 2 (switching the roles of left and right) shows that the car cannot reach Zillis if it is further apart from Argovia.

Case 4: $\ell \equiv r(\bmod 4)$. The car makes a net turn of $0^{\circ}$ after each section, so it must be facing east. We are going to show that, after traversing the first section, the car will be at $(1,0)$. It will be useful to interpret the Cartesian plane as the complex plane, i.e. writing $x+i y$ for $(x, y)$, where $i=\sqrt{-1}$. We will denote the $k$-th kilometer of movement by $m_{k-1}$,
which takes values from the set $\{1, i,-1,-i\}$, depending on the direction. We then just have to show that
$$\sum_{k=0}^{\ell r-1} m_{k}=1$$
which implies that the car will get to Zillis no matter how far it is apart from Argovia.

Case 4a: $\ell \equiv r \equiv 1(\bmod 4)$. First note that for $k=0,1, \ldots, \ell r-1$,
$$m_{k}=i^{\lfloor k / \ell\rfloor}(-i)^{\lfloor k / r\rfloor}$$
since $\lfloor k / \ell\rfloor$ and $\lfloor k / r\rfloor$ are the exact numbers of left and right turns before the $(k+1)$st kilometer, respectively. Let $a_{k}(\equiv k(\bmod \ell))$ and $b_{k}(\equiv k(\bmod r))$ be the remainders of $k$ when divided by $\ell$ and $r$, respectively. Then, since
$$a_{k}=k-\left\lfloor\frac{k}{\ell}\right\rfloor \ell \equiv k-\left\lfloor\frac{k}{\ell}\right\rfloor(\bmod 4) \quad \text{ and } \quad b_{k}=k-\left\lfloor\frac{k}{r}\right\rfloor r \equiv k-\left\lfloor\frac{k}{r}\right\rfloor(\bmod 4)$$
we have $\lfloor k / \ell\rfloor \equiv k-a_{k}(\bmod 4)$ and $\lfloor k / r\rfloor \equiv k-b_{k}(\bmod 4)$. We therefore have
$$m_{k}=i^{k-a_{k}}(-i)^{k-b_{k}}=\left(-i^{2}\right)^{k} i^{-a_{k}}(-i)^{-b_{k}}=(-i)^{a_{k}} i^{b_{k}}$$
As $\ell$ and $r$ are relatively prime, by Chinese Remainder Theorem, there is a bijection between pairs $\left(a_{k}, b_{k}\right)=(k(\bmod \ell), k(\bmod r))$ and the numbers $k=0,1,2, \ldots, \ell r-1$. Hence
$$\sum_{k=0}^{\ell r-1} m_{k}=\sum_{k=0}^{\ell r-1}(-i)^{a_{k}} i^{b_{k}}=\left(\sum_{k=0}^{\ell-1}(-i)^{a_{k}}\right)\left(\sum_{k=0}^{r-1} i^{b_{k}}\right)=1 \times 1=1$$
as required because $\ell \equiv r \equiv 1(\bmod 4)$.

Case 4b: $\ell \equiv r \equiv 3(\bmod 4)$. In this case, we get
$$m_{k}=i^{a_{k}}(-i)^{b_{k}}$$
where $a_{k}(\equiv k(\bmod \ell))$ and $b_{k}(\equiv k(\bmod r))$ for $k=0,1, \ldots, \ell r-1$. Then we can proceed analogously to Case 4a to obtain
$$\sum_{k=0}^{\ell r-1} m_{k}=\sum_{k=0}^{\ell r-1}(-i)^{a_{k}} i^{b_{k}}=\left(\sum_{k=0}^{\ell-1}(-i)^{a_{k}}\right)\left(\sum_{k=0}^{r-1} i^{b_{k}}\right)=i \times(-i)=1$$
as required because $\ell \equiv r \equiv 3(\bmod 4)$.

Now clearly the car traverses through all points between $(0,0)$ and $(1,0)$ during the first section and, in fact, covers all points between $(n-1,0)$ and $(n, 0)$ during the $n$-th section. Hence it will eventually reach $(d, 0)$ for any positive $d$.

To summarize: $(\ell, r)$ satisfies the required conditions if and only if
$$\ell \equiv r \equiv 1 \quad \text{ or } \quad \ell \equiv r \equiv 3 \quad(\bmod 4)$$