19. Let $A, B, C$ be points on the same plane with $\angle A C B=120^{\circ}$. There is a sequence of circles $\omega_{0}, \omega_{1}, \omega_{2}, \ldots$ on the same plane (with corresponding radii $r_{0}, r_{1}, r_{2}, \ldots$, where $r_{0}>r_{1}>r_{2}>\cdots$ ) such that each circle is tangent to both segments $C A$ and $C B$. Furthermore, $\omega_{i}$ is tangent to $\omega_{i-1}$ for all $i \geq 1$. If $r_{0}=3$, find the value of $r_{0}+r_{1}+r_{2}+\cdots$.
設 $A 、 B 、 C$ 為同一平面上的三點，其中 $\angle A C B=120^{\circ}$ 。現於同一平面上有一系列的圓 $\omega_{0} 、 \omega_{1} 、 \omega_{2} 、 \cdots$ (對應的半徑分別為 $r_{0} 、 r_{1} 、 r_{2} 、 \cdots$, 其中 $\left.r_{0}>r_{1}>r_{2}>\cdots\right)$, 當中每個均與線段 $C A$ 和 $C B$ 相切, 且對任意 $i \geq 1$ 皆有 $\omega_{i}$與 $\omega_{i-1}$ 相切。若 $r_{0}=3$, 求 $r_{0}+r_{1}+r_{2}+\cdots$ 的值。
19. $\frac{3}{2}+\sqrt{3}$
19. Consider two consecutive circles $\omega_{i}$ and $\omega_{i+1}$. Let their centres be $O_{i}$ and $O_{i+1}$ respectively. Since both circles are tangent to $C A$ and $C B$, both centres lie on the internal bisector of $\angle A C B$. In other words, $C, O_{i}, O_{i+1}$ are collinear and $\angle O_{i} C A=60^{\circ}$. Denote by $P_{i}$ and $P_{i+1}$ the points of tangency of $\omega_{i}$ and $\omega_{i+1}$ to $C A$ respectively. Then $O_{i} P_{i} \perp C A$ and $O_{i+1} P_{i+1} \perp C A$.

Consider the trapezium $O_{i} P_{i} P_{i+1} O_{i+1}$. We have
$O_{i} P_{i}=r_{i}, O_{i+1} P_{i+1}=r_{i+1}, O_{i} O_{i+1}=r_{i}+r_{i+1}$ and
$P_{i} P_{i+1}=P_{i} C-P_{i+1} C=\frac{1}{\sqrt{3}}\left(r_{i}-r_{i+1}\right)$. Note that
the last equality follows from $\angle O_{i} C P_{i}=60^{\circ}$.
Using Pythagoras' Theorem, we have
$$
\begin{aligned}
\left(\frac{1}{\sqrt{3}}\left(r_{i}-r_{i+1}\right)\right)^{2}+\left(r_{i}-r_{i+1}\right)^{2} & =\left(r_{i}+r_{i+1}\right)^{2} \\
\frac{2}{\sqrt{3}}\left(r_{i}-r_{i+1}\right) & =r_{i}+r_{i+1}
\end{aligned}
$$

This can be simplified to $r_{i+1}=(7-4 \sqrt{3}) r_{i}$. Therefore the radii of the circles form a geometric sequence with initial term 3 and common ratio is $7-4 \sqrt{3}$. It follows that the answer is
$$
\frac{3}{1-(7-4 \sqrt{3})}=\frac{3}{2}+\sqrt{3} \text {. }
$$