HMMT February

Solution:

Let $P$ satisfy $OP = x$. (For now, we focus on $f(P)$ and ignore the $A$ and $B$ from the problem statement.) The key idea is that if we invert at some point along $OP$ such that the images of $\Gamma$ and $\Omega$ are concentric, then $\omega_{i}$ still exist. Suppose that this inversion fixes $\Gamma$ and takes $\Omega$ to $\Omega'$ of radius $r$ (and $X$ to $X'$ in general). If the inversion is centered at a point $Q$ along ray $OP$ such that $OQ = d$, then the radius of inversion is $\sqrt{d^2-1}$. Let the diameter of $\Omega$ meet $OQ$ at $A$ and $B$ with $A$ closer to $Q$ than $B$. Then, $(AB; PP_{\infty}) = -1$ inverts to $(A'B'; P'Q) = -1$, where $P_{\infty}$ is the point at infinity along line $OP$, so $P'$ is the inverse of $Q$ in $\Omega'$. We can compute $OP' = \frac{r^2}{d}$ so $P'Q = d - \frac{r^2}{d}$ and $PQ = \frac{d^2-1}{d-\frac{r^2}{d}}$. Thus, we get the equation $\frac{d^2-1}{d-\frac{r^2}{d}} + x = d$, which rearranges to $\frac{1-r^2}{d^2-r^2} d = x$, or $d^2 - x^{-1}(1-r^2)d - r^2 = 0$.

Now, we note that the radius of $\Omega$ is

$$\frac{1}{2} AB = \frac{1}{2}\left(\frac{d^2-1}{d-r} - \frac{d^2-1}{d+r}\right) = \frac{r(d^2-1)}{d^2-r^2} = r\left(1 + \frac{r^2-1}{d^2-r^2}\right) = r\left(1 - \frac{x}{d}\right)$$

The quadratic formula gives us that $d = \frac{(1-r^2) \pm \sqrt{r^4 - (2-4x^2)r^2 + 1}}{2x}$, so $\frac{x}{d} = -\frac{1-r^2 \pm \sqrt{r^4 - (2-4x^2)r^2 + 1}}{2r^2}$, which means that the radius of $\Omega$ is

$$\frac{r^2+1 \pm \sqrt{r^4 - (2-4x^2)r^2 + 1}}{2r} = \frac{r + \frac{1}{r} \pm \sqrt{r^2 + \frac{1}{r^2} - 2 + 4x^2}}{2}$$

Note that if $r$ gives a valid chain of 2020 circles, so will $\frac{1}{r}$ by homothety/inversion. Thus, we can think of each pair of $r, \frac{1}{r}$ as giving rise to two possible values of the radius of $\Omega$, which are $\frac{r + \frac{1}{r} \pm \sqrt{r^2 + \frac{1}{r^2} - 1}}{2}$. This means that the pairs have the same sum of radii as the circles centered at $O$, and the product of the radii is $1-x^2$. (A simpler way to see this is to note that inversion at $P$ with radius $\sqrt{1-x^2}$ swaps the two circles.)

From this, it follows that the difference between the sum of the areas for each pair is $2\pi\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = \frac{5}{18}\pi$. There are $\frac{\varphi(2020)}{2} = 400$ such pairs, which can be explicitly computed as $\frac{1-\sin\frac{\pi k}{2020}}{1+\sin\frac{\pi k}{2020}}, \frac{1+\sin\frac{\pi k}{2020}}{1-\sin\frac{\pi k}{2020}}$ for positive integers $k < 1010$ relatively prime to $2020$.

Thus, the answer is $\frac{1000}{9}\pi$.