Tomi's Challenges - #18

The first step is to simplify everything as much as possible.

$\begin{cases} \cos a(\cos a+1)-1=2\sin b\cos c-\sin^2 a \\ \sin a=2\sin b\sin c+2021d^k+6k \\ d^k=\frac{a-2\cos b-6k}{2021} \\ \end{cases}$

$\begin{cases} \cos^2 a+\cos a+\sin^2 a-1=2\sin b\cos c \\ \sin a=2\sin b\sin c+2021d^k+6k \\ 2021d^k=a-2\cos b-6k \\ \end{cases}$

$\begin{cases} \cos a=2\sin b\cos c \\ \sin a=2\sin b\sin c+2021d^k+6k \\ a=2\cos b+2021d^k+6k \\ \end{cases}$

This system of equations is the intersection of the following parametric curve and surface.

$A=\begin{bmatrix} \cos a \\ \sin a \\ a \end{bmatrix}$ , $B=\begin{bmatrix} 2\sin b\cos c \\ 2\sin b\sin c+2021d^k+6k \\ 2\cos b+2021d^k+6k \end{bmatrix}$

The first vector represents a curve which is a helix around the origin whose radius is $1$ . The 2nd vector represents a sphere whose radius is $2$ and which is translated $2021d^k+6k$ units along the y and z axis. If $k$ is even, this number will always be positive and at least $6$ , and it will never intersect the helix. If $k$ is odd, this number could be anything and therefore there exist other parameters such that the sphere is intersected by the helix. Therefore $k$ can be $1, 3, 5, 7, 9$ .