Problem 6

For fixed $i$, the arcs belonging to different $j$ are pairwise disjoint: a tangent at the same point meeting both $C_j$ and $C_k$ would meet three circles. Let $\alpha_i$ be the total angular length of their union. Since each $j$ contributes two arcs of length $\theta_{ij}$, $\alpha_i=\sum_{j\ne i}2\theta_{ij}$. Now let $\beta_i$ be the angular length of the points on $C_i$ whose tangent is a supporting line of the union of all the closed disks. As the direction of a supporting line makes one full turn, its contact point runs through these support arcs, so $\sum_i\beta_i=2\pi$. The support arcs are disjoint from the pair-arcs. Therefore $\sum_i(\alpha_i+\beta_i)\le2n\pi$, and hence $\sum_i\alpha_i\le2(n-1)\pi$.