Finding points of self intersection

Question

A curve is parameterised by $$\mathbf{r}(t)=((2+\cos 3t)\cos 2t, (2+\cos 3t)(\sin 2t)), t\in [0,2\pi ]. $$ How can I find the self intersection points? I set $\mathbf{r}(t_1)=\mathbf{r}(t_2)$ and then tried to solve the simultaneous equations but it doesn't seem to work out. Any pointers?

Answer 1

Geometrically

Mathematica shows

Three point are $$(1,1.7),(-2,0),(1,-1.7)$$

Analytically

(deleted)

Answer 2

Parametric representation

$$\mathbf{r}(t)=((2+\cos 3t)\cos 2t, (2+\cos 3t)\sin 2t)), t\in [0,2\pi ]$$

can be replaced by this (complex) one

$$\mathbf{r}(t)=(2+\cos 3t)e^{2it} , t\in [0,2\pi ]$$

Thus, we are looking for values of $t_1$ and $t_2$, $t_1 \neq t_2$, such that:

$$\tag{1}(2+\cos 3t_1)e^{2it_1}=(2+\cos 3t_2)e^{2it_2}$$

Two complex numbers are equal if and only if their modules are equal and their arguments are equal (modulo $2 \pi$).

Remark: Due to the fact that $2+\cos 3t>0$ for any value of $t$ shows that $2+\cos 3t_1$ and $2+\cos 3t_2$ are the modules of the LHS and RHS of (1), resp.

Thus, (1) is equivalent to:

$$\cases{\cos(3t_1)=\cos(3 t_2)\\2t_1=2t_2 \ modulo \ 2 \pi}$$

As $$cos(u)=cos(v) \ \iff \ u=\pm v+K 2\pi:$$

the previous conditions are equivalent to

$$\cases{3t_1= s 3 t_2+K 2\pi\\2t_1=2t_2+K' 2 \pi}$$

where $s=\pm1$ and $K,K'$ are integers.

$$\tag{2}\iff \ \ \cases{t_1= s t_2+K 2\pi/3\\t_1=t_2+K' \pi}$$

The cases where $s=1$ will not give double points (because they lead to $t_1=t_2+K''2\pi$). We can thus assume $s=-1$, i.e.,

$$\tag{3}\iff \ \ \cases{t_1= -t_2+K 2\pi/3 \ \ (a)\\t_1=t_2+K' \pi \ \ (b)}$$

Adding and substracting (3)(a) and (3)(b):

$$\tag{4}\cases{2t_1=K2\pi/3+K'\pi \ \ \iff \ \ t_1=K\pi/3+K'\pi/2\ \ (a)\\2t_2=K2\pi/3-K'\pi \ \ \iff \ \ t_2=K\pi/3-K'\pi/2 \ \ (b)}$$

with the same values of $K$ and $K'$ in (4)(a) and (4)(b).

For example, if $K=2$ and $K'=1$, $(t_1,t_2)=(7\pi/6,\pi/6)$, giving point $(1,\sqrt{3}).$

I leave you the task to consider the different other cases (by taking different possible cases for integers $K$ and $K'$, some of them being redundant). You will end up with the three following solutions (up to an exchange between $t_1$ and $t_2$, of course):

$$(t_1,t_2)=(7\pi/6,\pi/6), \ \ (5\pi/6,11\pi/6), \ \ \ \ (\pi/2,3\pi/2), $$ yielding double points :

$$(1,\sqrt{3}), \ \ (1,-\sqrt{3}), \ \ (-2,0)$$

respectively.