Let $A O B$ be a 60 -degree angle. For any point $P$ in the interior of $\angle A O B$, let $A^{\prime}$ and $B^{\prime}$ be the feet of the perpendiculars from $P$ to $A O$ and $B O$ respectively. Denote by $r$ and $s$ the distances $O P$ and $A^{\prime} B^{\prime}$. Find all possible pairs of real numbers $(r, s)$.
Extend $A^{\prime} P$ to meet $O B$ at $Z$. Notice that, because $\angle O A^{\prime} P$ and $\angle O B^{\prime} P$ are both right, the circle with diameter $O P$ passes through $O, P, A^{\prime}$, and $B^{\prime}$. Thus $\angle B^{\prime} O P=$ $\angle B^{\prime} A^{\prime} P$ since both intercept the same arc on this circle, and $\triangle Z O P \sim \triangle Z A^{\prime} B^{\prime}$ by AA. We get

$$
\frac{s}{r}=\frac{B^{\prime} A^{\prime}}{O P}=\frac{Z A^{\prime}}{Z O}=\frac{\sqrt{3}}{2}
$$

because $Z O A^{\prime}$ is a $30-60-90$ triangle. Since $r$ can obviously take on any value, the possibilities for $(r, s)$ are $\left(r, \frac{r \sqrt{3}}{2}\right)$ for

![](https://cdn.mathpix.com/cropped/2024_04_17_2fae5e31928f586282e5g-2.jpg?height=423&width=287&top_left_y=274&top_left_x=1060)
every positive real $r$.