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Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Consider a quarter-circle with center $O$ , arc $\widehat{A B}$ , and radius $2$ . Draw a semicircle with diameter $\overline{O A}$ lying inside the quarter-circle. Points $P$ and $Q$ lie on the semicircle and segment $\overline{O B}$ , respectively, such that line $P Q$ is tangent to the semicircle. As $P$ and $Q$ vary, compute the maximum possible area of triangle $B Q P$ .

Proposed by: Daeho Jacob Lee

Answer: $\frac{1}{2} = 0.5$

Solution

Solution:

Note that we can bound the area of $\triangle B Q P$ by
$\begin{aligned} {[B Q P]} & = \frac{1}{2} B Q \cdot Q P \sin \angle B Q P \\ & \leq \frac{1}{2} B Q \cdot Q P \\ & = \frac{1}{2} B Q (2 - B Q) \\ & \leq \frac{1}{2} . \end{aligned}$

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