Two moving points from the origin

$$ $P$ and $Q$ will meet again if $y_P=y_Q$ and $t_P=t_Q$ . We have $v_P = \frac{dy_P}{dt} = \sin \pi t$ and $v_Q = \frac{dy_Q}{dt} = 2\sin 2\pi t$ , then $$ $\begin{aligned} y_P &=\int_0^{\frac{a}{b}} \sin \pi t\, dt\\ &=-\left.\frac{\cos \pi t}{\pi}\right|_0^{\frac{a}{b}}\\ &= \frac{1- \cos \frac{a}{b} \pi}{\pi} \end{aligned}$ and $\begin{aligned} y_Q &=2\int_0^{\frac{a}{b}} \sin 2\pi t\, dt\\ &=-\left.\frac{\cos 2\pi t}{\pi}\right|_0^{\frac{a}{b}}\\ &= \frac{1- \cos \frac{2a}{b} \pi}{\pi}. \end{aligned}$ Therefore $\begin{aligned} y_P &=y_Q\\ \frac{1- \cos \frac{a}{b} \pi}{\pi}&= \frac{1- \cos \frac{2a}{b} \pi}{\pi}\\ \cos \frac{2a}{b} \pi-\cos \frac{a}{b} \pi&=0\\ 2\cos^2 \frac{a}{b} \pi-\cos \frac{a}{b} \pi - 1&=0\\ \left(\cos \frac{a}{b} \pi - 1\right)\left(2\cos \frac{a}{b} \pi + 1\right)&=0. \end{aligned}$ We obtain $$ $\begin{aligned} \cos \frac{a}{b} \pi - 1&=0\\ \cos \frac{a}{b} \pi&=1\\ \frac{a}{b}\pi&=2\pi n_1 & \text{ for } n_1 \in \mathbb{Z,}\\ \frac{a}{b}&=2 n_1\\ \\ 2\cos \frac{a}{b} \pi + 1&=0\\ \cos \frac{a}{b} \pi &=-\frac{1}{2}\\ \frac{a}{b} \pi &=\frac{2}{3} \pi +2\pi n_2 \,\color{#3D99F6}{\text{or}}\, \frac{a}{b} \pi =\frac{4}{3} \pi +2\pi n_3 & \text{ for } n_2 \text{ and } n_3\in \mathbb{Z,}\\ \frac{a}{b}&=\frac{2}{3} +2 n_2 \,\color{#3D99F6}{\text{or}}\, \frac{a}{b}=\frac{4}{3} +2 n_3. \end{aligned}$ $P$ and $Q$ will meet for the first time if $n_2=0$ . Thus, $\frac{a}{b}=\frac{2}{3}$ and $a+b=\boxed{5}$ . $$

$\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$