Problem 1161

4. Let the center of circle $C_{1}$ be $O$, and the radius be $r$. Connect $O A_{i}, O B_{i} (i=1,2, \cdots, n)$. Applying Ptolemy's inequality in quadrilateral $O A_{1} B_{1} B_{2}$, we have $O A_{1} \cdot B_{1} B_{2} + O O_{2} \cdot A_{1} B_{1} \geqslant O B_{1} \cdot A_{1} B_{2}$, which means $r \cdot B_{1} B_{2} + \lambda r \cdot A_{1} B_{1} \geqslant \lambda r (A_{1} A_{2} + A_{2} B_{2})$, hence
$$B_{1} B_{2} + \lambda A_{1} B_{1} \geqslant \lambda (A_{1} A_{2} + A_{2} B_{2}).$$

Similarly, applying Ptolemy's inequality iteratively, we have $B_{2} B_{3} + \lambda' A_{2} B_{2} \geqslant \lambda (A_{2} A_{3} + A_{3} B_{3})$; $B_{3} B_{4} + \lambda \cdot A_{3} B_{3} \geqslant \lambda (A_{3} A_{4} + A_{4} B_{4})$; $\cdots$; $B_{n-1} B_{n} + \lambda \cdot A_{n-1} B_{n-1} \geqslant \lambda (A_{n-1} A_{n} + A_{n} B_{n})$; $B_{n} B_{1} + \lambda A_{n} B_{n} \geqslant \lambda (A_{n} A_{1} + A_{1} B_{1})$.

Adding these inequalities, we get $B_{1} B_{2} + B_{2} B_{3} + \cdots + B_{n-1} B_{n} + B_{n} B_{1} \geqslant \lambda (A_{1} A_{2} + A_{2} A_{3} + \cdots + A_{n-1} A_{n} + A_{n} A_{1})$, hence $p_{2} \geqslant \lambda p_{1}$. The equality in Ptolemy's inequality holds if and only if the quadrilaterals $O A_{1} B_{1} B_{2}$, $O A_{2} B_{2} B_{3}$, $\cdots$, $O A_{n} B_{n} B_{1}$ are all cyclic quadrilaterals. By the properties of cyclic quadrilaterals, we know $\angle O A_{2} A_{3} = \angle O B_{2} B_{3}$, $\angle O A_{2} A_{1} = \angle O B_{3} B_{2}$, but $\angle O B_{2} B_{3} = \angle O B_{3} O_{2}$, thus $\angle O A_{2} A_{1} = \angle O A_{2} A_{3}$, so $\triangle O A_{1} A_{2} \cong \triangle O A_{2} A_{3}$, therefore, $A_{1} A_{2} = A_{2} A_{3}$. Similarly, $A_{2} A_{3} = A_{3} A_{4} = \cdots = A_{n} A_{1}$, which means the $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ is a regular $n$-sided polygon.

Conversely, if $A_{1} A_{2} \cdots A_{n}$ is a regular $n$-sided polygon, rotating it counterclockwise around point $O$ by $\frac{2 \pi}{n}$, we have $A_{1} \rightarrow A_{2}$, $A_{2} \rightarrow A_{3}$, $\cdots$, $A_{n} \rightarrow A_{1}$, thus $B_{1} \rightarrow B_{2}$, $B_{2} \rightarrow B_{3}$, $\cdots$, $B_{n} \rightarrow B_{1}$. Therefore, $B_{1} B_{2} \cdots B_{n}$ is also a regular $n$-sided polygon, so $A_{1} A_{2} = A_{2} A_{3} = \cdots = A_{n} A_{1} = 2 r \cdot \sin \frac{\pi}{n}$, $B_{1} B_{2} = B_{2} B_{3} = \cdots = B_{n} B_{1} = 2 \lambda r \cdot \sin \frac{\pi}{n}$. At this point, we have $p_{2} = \lambda p_{1}$.