Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Let $A O B$ be a $60$ -degree angle. For any point $P$ in the interior of $\angle A O B$ , let $A'$ and $B'$ 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 $A' B'$ . Find all possible pairs of real numbers $(r, s)$ .

Solution

Solution:

Extend $A' P$ to meet $O B$ at $Z$ . Notice that, because $\angle O A' P$ and $\angle O B' P$ are both right, the circle with diameter $O P$ passes through $O, P, A'$ , and $B'$ . Thus $\angle B' O P = \angle B' A' P$ since both intercept the same arc on this circle, and $\triangle Z O P \sim \triangle Z A' B'$ by AA. We get
$\frac{s}{r} = \frac{B' A'}{O P} = \frac{Z A'}{Z O} = \frac{\sqrt{3}}{2}$
because $Z O A'$ 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

every positive real $r$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.