Rosy Polar Curve

We write $k$ as $\frac{p}{q}$ , where $p$ and $q$ are positive, relatively prime integers. The curve begins ( $\theta=0$ ) at $r=1$ on the polar axis, or equivalently the point $(1,0)$ on the $x$ -axis. It reaches the pole (the origin) for the first time when $\frac{p}{q}\theta = \frac{\pi}{2}$ , i.e. when $\theta = \frac{q\pi}{2p}$ . By then it has completed half of a petal, which means that $\theta$ has to rotate through $\frac{q\pi}{p}$ to complete one full petal.

In order to complete the rose, the curve must return to its starting point, which will happen in one of two ways: either $\theta$ is a multiple of $2\pi$ and $r=1$ , the latter implying that $\frac{p}{q}\theta$ is also a multiple of $2\pi$ ; or $\theta$ is an odd multiple of $\pi$ and $r=-1$ , the latter again implying that $\frac{p}{q}\theta$ is also an odd multiple of $\pi$ .

Given that $p$ and $q$ are relatively prime, we now have to consider three cases: $p$ is odd and $q$ is even (henceforth referred to as $\frac{O}{E}$ ); $p$ is even and $q$ is odd ( $\frac{E}{O}$ ); and $p$ and $q$ are both odd ( $\frac{O}{O}$ ).

Note that in the first two cases, $\frac{O}{E}$ and $\frac{E}{O}$ , it is not possible for $\theta$ and $\frac{p}{q}\theta$ to both be odd multiples of $\pi$ ; they can however both be multiples of $2\pi$ , and the smallest positive $\theta$ for which this will happen is $2q\pi$ . This means that, since one petal requires $\theta$ to rotate through $\frac{q\pi}{p}$ , a complete rose will consist of $2p$ petals.

On the other hand, in the third case, ( $\frac{O}{O}$ ), it is not possible for $\theta$ and $\frac{p}{q}\theta$ to both be multiples of $2\pi$ ; they can however both be odd multiples of $\pi$ , and the smallest positive $\theta$ for which this will happen is $q\pi$ . In this case, a complete rose will only have $p$ petals.

We are now ready to count the number of times a rose will intersect one or both of the axes. This happens whenever $\theta$ is a multiple of $\frac{\pi}{2}$ , or when $\frac{p}{q}\theta$ is an odd multiple of $\frac{\pi}{2}$ , the latter because then $r=0$ and the curve will go through the pole. However, because these two conditions can potentially occur for the same value of $\theta$ , we must be careful to remove any "overlaps".

Case ( $\frac{O}{E}$ ): Since $\theta$ must rotate through $2q\pi$ to complete the rose, it will be a multiple of $\frac{\pi}{2}$ a total of $4q$ times. Since this rose will have $2p$ petals, it will go through the pole $2p$ times. Can these happen simultaneously? Suppose $\frac{p}{q}\theta$ is an odd multiple of $\frac{\pi}{2}$ , i.e. $\frac{p}{q}\theta\ = \frac{(2n+1)\pi}{2}$ for some whole $n$ ; then $\theta = \frac{(2n+1)q\pi}{2p}$ . For $\theta$ to be a multiple of $\frac{\pi}{2}$ , $(2n+1)$ would have to be a multiple (and by necessity, an odd multiple) of $p$ ; this will only happen twice within the range $0 \le \theta < 2q\pi$ , namely when $(2n+1) = p$ and $(2n+1) = 3p$ . Thus the number of times this rose will intersect the axes is $\color{#3D99F6}4q+2p-2$ .

Case ( $\frac{E}{O}$ ): By the same reasoning as in the above case, $\theta$ will be a multiple of $\frac{\pi}{2}$ a total of $4q$ times and the rose will go through the pole $2p$ times. Again, to see whether these can happen simultaneously, we let $\frac{p}{q}\theta\ = \frac{(2n+1)\pi}{2}$ for some whole $n$ , and again, $\theta = \frac{(2n+1)q\pi}{2p}$ . This time, however, $\theta$ can never be a multiple of $\frac{\pi}{2}$ as $(2n+1)$ and $q$ are both odd while $p$ is even. So the number of times this rose will intersect the axes is $\color{#3D99F6}4q+2p$ .

Case ( $\frac{O}{O}$ ): Here $\theta$ must rotate through $q\pi$ to complete the rose, so it will be a multiple of $\frac{\pi}{2}$ only $2q$ times. Also, since this rose will only have $p$ petals, it will go through the pole only $p$ times. Once again to check for overlap, we let $\frac{p}{q}\theta\ = \frac{(2n+1)\pi}{2}$ for some whole $n$ , and once more, $\theta = \frac{(2n+1)q\pi}{2p}$ . This will only be a multiple of $\frac{\pi}{2}$ once within the range $0 \le \theta < q\pi$ , namely when $(2n+1) = p$ . Thus the number of times this rose will intersect the axes is $\color{#3D99F6}2q+p-1$ .

Finally, we can check whether any of the three cases can have exactly 2018 axes crossings.

Case ( $\frac{O}{E}$ ): We get $4q+2p-2=2018$ , i.e $2q+p=1010$ , which has no valid solutions as $p$ is odd.

Case ( $\frac{E}{O}$ ): We get $4q+2p=2018$ , i.e $2q+p=1009$ , which has no valid solutions as $p$ is even.

Case ( $\frac{O}{O}$ ): We get $2q+p-1=2018$ , i.e $2q+p=2019$ , which has valid solutions.

The equation $2q+p=2019$ has 1009 solutions over positive integers, from $(p,q) = (1,1009)$ to $(p,q) = (2017,1)$ . We need $p$ and $q$ to both be odd; for this it suffices that $q$ is odd. We also need to ensure that $p$ and $q$ are relatively prime. As $2019$ has prime factors $3$ and $673$ , we need $p$ and $q$ to not both be multiples of one of those two numbers; the former would happen if $q$ is a multiple of three, the latter would only happen in one case, if $q = p = 673$ .

To ensure $q$ is neither even nor a multiple of three, we use inclusion-exclusion; from 1009, we subtract the $\left \lfloor \frac{1009}{2} \right \rfloor$ multiples of $2$ as well as the $\left \lfloor \frac{1009}{3} \right \rfloor$ multiples of $3$ , and then add back the $\left \lfloor \frac{1009}{6} \right \rfloor$ multiples of $6$ . Finally we subtract $1$ for the single case $(673,673)$ .

Thus the number of unique $k$ 's for which the rose $r = \cos(k\theta)$ intersects the axes exactly 2018 times is

$1009 - \left \lfloor \frac{1009}{2} \right \rfloor - \left \lfloor \frac{1009}{3} \right \rfloor\ + \left \lfloor \frac{1009}{6} \right \rfloor -1 = 1009 - 504 - 336 + 168 -1 = \boxed{336}$