Rose Petals

The curve $r = \cos (k \theta)$ 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 $k \theta = \frac{\pi}{2}$ , i.e. when $\theta = \frac{\pi}{2k}$ . By then it has completed half of a petal, which means that $\theta$ has to rotate through $\frac{\pi}{k}$ to complete one full petal.

In order to complete a 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 $k \theta$ is also a multiple of $2\pi$ ; or $\theta$ is an odd multiple of $\pi$ and $r=-1$ , the latter again implying that $k \theta$ is also an odd multiple of $\pi$ .

Suppose $k$ is odd. The first time one of the two above conditions is satisfied is when $\theta = \pi$ ; by then, since one petal is completed for every $\frac{\pi}{k}$ , the rose will have $k$ petals. But this means the number of petals will be odd.

Suppose $k$ is even. Then when $\theta$ is any multiple of $\pi$ (even an odd multiple) $k \theta$ will be an even multiple of $\pi$ , so the first time the necessary conditions for completion of a rose are satisfied will be when $\theta = 2 \pi$ ; by then, the rose will have $2k$ petals. But this means the number of petals will be a multiple of $4$ .

Since $2018$ is neither odd nor a multiple of $4$ , there can be no integer $k$ such that $P(k) = 2018$ .

If $r = \cos{c\theta}$ , for some positive integer $c$ , then the number of petals will be $2c$ if $c$ is even, or the number of petals will be $c$ if $c$ is odd. Therefore, the number of petals must be odd or a multiple of $4$ , but $2018$ is neither.

Prime factorization of 2018 is 2 and 1009. It (2018=2×1009) will not match any odd or even (multiple of 4) petals counts in a polar plot r=cos(k*theta), for any value of k (odd or even).

Therefore, answer is 'NO'.

Since if $k$ is odd,then $P(k)=k$ .If $k$ is even,then $P(k)$ is divisible by 4.Since 2018 isn't odd or divisible by 4,so the answer is $\boxed{No}$

roses are red, violets are blue, 1009 is odd so saying there is an integer k such that P(k)=2018 won't be true.

We have to use facts.....such that no of petals in a flower appears to be in Fibonacci sequence....and we just have to prove that 2018 is not a Fibonacci number

2018 is not a multiple of 4 or odd.

P (odd) = odd , P (even) = 2 × even

Why?

We will count the number of petals by counting how many times $r=1$ or $r=-1$ (so we will count the endpoints of petals)

At first notice that the period of $cos (kθ)$ is equal to $\frac {2π}{k}$ .

So if we segmented $[0, 2π$ ) into pieces $[0+n\frac {2π}{k}, \frac {2π}{k} + n\frac {2π}{k})$ we would get in each of $k$ pieces 1 solution ( $θ=0+n\frac {2π}{k}$ ) for $cos (kθ)=1$ and 1 solution ( $θ=\frac {π}{k}+n\frac {2π}{k}$ ) for $cos (kθ)=-1$ .

The only issue when $θ=\frac {π}{k}+n\frac {2π}{k}, r=-1$ is that we might count it twice (if $θ=\frac {π}{k}+n\frac {2π}{k}+π \Rightarrow r=1$ ).

This happens when k is odd and doesn't when k is even.

Since 2018 is not divisible by 4 nor odd there is no such k that $P (k) = 2018$

• Hello! I found an interesting solution for this problem. Maybe not so elegant as most of the others found here. First we must divide the circle in 4k sectors. If we follow how the plot is created, as angle theta is varying from 0 to 360 deg, we will observe that it starts in sector 1, after that it jumps in sector 1+(2k+1), after this in sector 1+(2k+1)+1, after this in sector 1+(2k+1)+1 +(2k+1) and so on. Also, we can observe, that the condition to obtain a closed curve is that at some step we will obtain the rest 1 when we divide the aforementioned sum at 4k.That step is equal to 2 P(k)+1. So this sum, at the step 2 P(k)+1, should be equal with some constant multiplied with 4k, plus 1. If we write this we find that 2 * 2018 * (k+1)+1 = 4 C k+1. Finally we obtain that 1009=(C-1009)*k which has no solution because min(k)=P(k) and C-1009 > 0. THE ANSWER IS NO! :D

r=cos(k∆),if k is an even integer then 2k should be the number of petals p(k),if k is an odd integer then k should be the p(k).But the number of petals in the flowers depends on fibonicci series:0,1,1,2,3,5.....So here p(k)=2018,but 2018 is not a fibonicci number,so the answer is "no"