In the figure, $O$ is the center of the semicircle with diameter $P Q, R$ is a point on the semicircle, and $R M$ is perpendicular to $P Q$. If the measure of arc $\widehat{P R}$ is twice the measure of arc $\overparen{R Q}$, what is the ratio between $P M$ and $M Q$?

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-026.jpg?height=249&width=371&top_left_y=2297&top_left_x=1462)
If the arc $\widehat{P R}$ is twice the arc $\overparen{R Q}$, the same relationship holds for the central angles, that is, $P \widehat{O} R=2 R \widehat{O} Q$. Since $P \widehat{O} R+R \widehat{O} Q=180^{\circ}$, it follows that

$$
180^{\circ}=2 R \widehat{O} Q+R \widehat{O} Q=3 R \widehat{O} Q
$$

from which $R \widehat{O} Q=60^{\circ}$. But, $O R=O Q$ is the radius of the circle, so the triangle $\triangle O R Q$ is equilateral. Thus, its height $R M$ is also the median, that is, $O M=M Q$. If $r$ is the radius of the circle, then $O M=M Q=\frac{1}{2} r$ and

$$
\frac{P M}{M Q}=\frac{P O+O M}{M Q}=\frac{r+\frac{1}{2} r}{\frac{1}{2} r}=3
$$

that is, the ratio between $P M$ and $M Q$ is 3.