Tomi's Challenges - #9

When you have a parametric curve, it's also common to describe it using a vector:

$\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} 2020\cos (\alpha+\beta)\cos \gamma-2021\sin(\alpha+\beta)\sin\gamma\\ 2020\sin (\alpha+\beta)\cos \gamma+2021\cos(\alpha+\beta)\sin\gamma \end{bmatrix}$

Without loss of generality, let $\theta=\alpha+\beta$ .

$\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} 2020\cos (\theta)\cos \gamma-2021\sin(\theta)\sin\gamma\\ 2020\sin (\theta)\cos \gamma+2021\cos(\theta)\sin\gamma \end{bmatrix}$

Notice that this vector is of the form

$\begin{bmatrix} x\cos(t)-y\sin(t)\\ x\sin(t)+y\cos(t) \end{bmatrix}=\begin{bmatrix} \cos(t) & -\sin(t)\\ \sin(t) & \cos(t) \end{bmatrix}\ \begin{bmatrix} x\\ y \end{bmatrix}$

Rewriting our vector, we get:

$\begin{bmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{bmatrix}\ \begin{bmatrix} 2020\cos(\gamma)\\ 2021\sin(\gamma) \end{bmatrix}$

We can now see that this parametric curve is nothing but an ellipse with the shorter radius being $2020$ , the longer one being $2021$ , which is being rotated by an arbitrary $\theta$ . By rotating it around its origin, we get a ring as the union of all points which the ellipse goes through, with its inner radius being $2020$ and width being $2021-2020=1$ . If we circumscribe a square around this ring, we are left to compute the complement of the area of the ring.

Here is a visualization:

So the area is $P=2020^2\pi+4042^2-2021^2\pi=4042^2+(2020-2021)(2020+2021)\pi=4042^2-4041\pi$ . After computing $P$ , we get $\boxed{16325068.82}$ .