Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

For each disc $D$ in $\mathcal{D}$, let $\omega_D$ denote the centre of $D$, and let $\alpha_D$ be the arc-length of the image of $D$ under radial projection from $O$ onto the unit circle centred at $O$. Clearly, $\alpha_D/2 > \sin(\alpha_D/2) = 1/O\omega_D$.

Now, for each positive integer $k < n$, let $\mathcal{D}_k$ be the set of those discs in $\mathcal{D}$ whose centres lie in the closed disc of radius $k+1$ centred at $O$. Since $\mathcal{D}_i \subseteq \mathcal{D}_j$ if $i \le j$, and each $\mathcal{D}_k$ contains at least $k$ elements, we may recursively choose (or apply Hall's marriage theorem to produce) a system of distinct representatives, $D_1, \dots, D_{n-1}$, for the collection $\mathcal{D}_1, \dots, \mathcal{D}_{n-1}$, to obtain
$$\sum_{D \in \mathcal{D}_{n-1}} \alpha_D > 2 \sum_{D \in \mathcal{D}_{n-1}} 1/O\omega_D \ge 2 \sum_{k=1}^{n-1} 1/O\omega_{D_k} \ge 2 \sum_{k=1}^{n-1} 1/(k+1) > 2 \log \frac{n+1}{2}.$$
Finally, if $N_{n-1}$ is the maximal number of discs in $\mathcal{D}_{n-1}$ stabbed by a line through $O$ as it performs a half-turn about $O$, then $\pi N_{n-1} \ge \sum_{D \in \mathcal{D}_{n-1}} \alpha_D$ and the conclusion follows.