HMMT November, 2023

Solution:

The key claim is that $\mathbb{E}\left[XY^{2}\right] = d^{2} + r_{1}^{2} + r_{2}^{2}$.

To prove this claim, choose an arbitrary point $B$ on $\omega_{2}$. Let $r_{1}, r_{2}$ be the radii of $\omega_{1}, \omega_{2}$ respectively, and $O_{1}, O_{2}$ be the centers of $\omega_{1}, \omega_{2}$ respectively. Thus, by the law of cosines, $\overline{O_{1}B} = \sqrt{d^{2} + r_{2}^{2} - 2 r_{2} d \cos(\theta)}$, where $\theta = \angle O_{1}O_{2}B$. Since the average value of $\cos(\theta)$ is $0$, the average value of $\left(\overline{O_{1}B}\right)^{2}$ is $d^{2} + r_{2}^{2}$.

Now suppose $A$ is an arbitrary point on $\omega_{1}$. By the law of cosines, $\overline{AB}^{2} = \left(\overline{O_{1}B}\right)^{2} + r_{1}^{2} - 2 r_{1} d \cos(\theta)$, where $\theta = \angle AO_{1}B$. Thus, the expected value of $\overline{AB}^{2}$ is the expected value of $\left(\overline{O_{1}B}\right)^{2} + r_{1}^{2}$, which becomes $d^{2} + r_{1}^{2} + r_{2}^{2}$. This proves the key claim.

Thus, we have $d^{2} + r_{1}^{2} + r_{2}^{2} = 2023$. The lengths of the internal and the external tangents give us $d^{2} - (r_{1} + r_{2})^{2} = 361$, and $d^{2} - (r_{1} - r_{2})^{2} = 1369$. Thus,
$$d^{2} - (r_{1}^{2} + r_{2}^{2}) = \frac{\left(d^{2} - (r_{1} + r_{2})^{2}\right) + \left(d^{2} - (r_{1} - r_{2})^{2}\right)}{2} = \frac{361 + 1369}{2} = 865$$
Thus, $d^{2} = \frac{865 + 2023}{2} = 1444 \Longrightarrow d = 38$.

Solution:

We present another way of showing that $\mathbb{E}\left[XY^{2}\right] = d^{2} + r_{1}^{2} + r_{2}^{2}$ using complex numbers. The finish is the same as Solution 1.

Let the center of $\omega_{1}$ and $\omega_{2}$ be $0$ and $k$, respectively. Select $Z_{1}$ and $Z_{2}$ uniformly random on unit circle. Then,
$$\begin{aligned} \mathbb{E}\left[XY^{2}\right] &= \mathbb{E}\left|k + r_{1} Z_{1} + r_{2} Z_{2}\right|^{2} \\ &= \mathbb{E}\left(k + r_{1} Z_{1} + r_{2} Z_{2}\right)\left(\bar{k} + r_{1} \overline{Z_{1}} + r_{2} \overline{Z_{2}}\right) \end{aligned}$$
Then, observe that
$$\mathbb{E}\left[Z_{1}\right] = \mathbb{E}\left[Z_{2}\right] = \mathbb{E}\left[Z_{1} \overline{Z_{2}}\right] = \mathbb{E}\left[Z_{2} \overline{Z_{1}}\right] = 0$$
so when expanding, six terms vanish, leaving only
$$\mathbb{E}\left[XY^{2}\right] = \mathbb{E}\left[k \bar{k} + r_{1}^{2} Z_{1} \overline{Z_{1}} + r_{2}^{2} Z_{2} \overline{Z_{2}}\right] = d^{2} + r_{1}^{2} + r_{2}^{2}$$
The rest of the computation is as in Solution 1, so $d = 38$.