As shown in Figure 1, quadrilateral $A O B C$ is a rectangle, with $A(0,2)$ and $B(4,0)$. Points $P$ and $Q$ start moving from point $O$ simultaneously. Point $P$ moves along the path $O A C B$, and point $Q$ moves along the path $O B C A$. $P Q$ intersects $O C$ at $R$. Point $P$ moves at a speed of 3 units per second, and point $Q$ moves at a speed of 2 units per second, stopping when they meet. Let the time of movement be $t$ seconds. For what value of $t$ are $\triangle A R O$ and $\triangle R C B$ similar?
Solution: If $\triangle A R O \sim \triangle R C B$, then
$\angle A O C=\angle O C B, A O=B C=2$.
Thus, $\frac{A O}{R C}=\frac{O R}{B C}$, which means $\frac{2}{R C}=\frac{O R}{2}$.
Hence, $R C \cdot O R=4$.
Also, $O C=\sqrt{2^{2}+4^{2}}=2 \sqrt{5}$,
which means $R C+O R=2 \sqrt{5}$.
From equations (1), (2), and $O R_{1}=\sqrt{5}-1, O R_{2}=\sqrt{5}+1$,
it follows that when $O R_{1}=\sqrt{5}-1, R_{1} C=\sqrt{5}+1$ or $O R_{2}=\sqrt{5}+1$, $R_{2} C=\sqrt{5}-1$, $\triangle A R O \sim \triangle R C B$;
Additionally, when $R_{0}$ is the midpoint of $O C$, the two triangles are also similar.
$1^{\circ}$ When $0<t \leq \frac{2}{3}$, $O R<\sqrt{5}-1$, at this time the two triangles are not similar.
When $t=\frac{2}{3}$, we have $P(0,2), Q\left(\frac{4}{3}, 0\right)$.
The equation of line $P Q$ is $y=-\frac{3}{2} x+2$,
and the equation of line $O C$ is $y=\frac{1}{2} x$.
From equations (3) and (4), we get $R\left(1, \frac{1}{2}\right)$, at this time
$O R=\sqrt{1^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{5}}{2}<\sqrt{5}-1$.
$2^{\circ}$ When $\frac{2}{3}<t<2$,
if $\frac{O Q}{P C}=\frac{O R_{0}}{R_{0} C} \Rightarrow \frac{2 t}{6-3 t}=\frac{\sqrt{5}}{\sqrt{5}} \Rightarrow t=\frac{6}{5}$;
if $\frac{O Q}{P C}=\frac{O R_{1}}{R_{1} C} \Rightarrow \frac{2 t}{6-3 t}=\frac{\sqrt{5}-1}{\sqrt{5}+1} \Rightarrow t=\frac{6(6-\sqrt{5})}{31}$;
if $\frac{O Q}{P C}=\frac{O R_{2}}{R_{2} C} \Rightarrow \frac{2 t}{6-3 t}=\frac{\sqrt{5}+1}{\sqrt{5}-1} \Rightarrow t=\frac{6(6+\sqrt{5})}{31}$.
$3^{\circ}$ When $2 \leq t \leq \frac{12}{5}$, $P$ and $Q$ are both on $B C$, and $\triangle R C B$ does not

exist.
Therefore, when $t=\frac{6(6-\sqrt{5})}{31}$ or $t=\frac{6(6+\sqrt{5})}{31}$ or $t=\frac{6}{5}$, $\triangle A R O \sim \triangle R C B$.
(Xie Anjia, Zhongxiang No.2 High School, Hubei Province, 431900)