Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.4 National Olympiad Prove it Philippines

Problem:
On the line containing diameter $AB$ of a circle, a point $P$ is chosen outside of this circle, with $P$ closer to $A$ than $B$ . One of the two tangent lines through $P$ is drawn. Let $D$ and $E$ be two points on the tangent line such that $AD$ and $BE$ are perpendicular to it. If $DE = 6$ , find the area of triangle $BEP$ .

Solution

Solution:
$FB = \sqrt{AB^{2} - AF^{2}} = \sqrt{64 - 36} = 2\sqrt{7}$

The radius $OC$ is the midsegment of trapezoid $ADED$ . Hence $[ABED] = 4 \times 6 = 24$ . Moreover
$\begin{aligned} [ABED] & = [ABF] + [ADEF] \\ 24 & = \frac{2\sqrt{7} \times 6}{2} + AD \times 6 \\ AD & = 4 - \sqrt{7} \\ EB & = AD + FB = 4 + \sqrt{7} \end{aligned}$
Also by similarity (triangle $PAD$ and triangle $PBE$ )
$\begin{aligned} \frac{PD}{AD} & = \frac{PE}{EB} \\ \frac{PD}{4 - \sqrt{7}} & = \frac{PD + 6}{4 + \sqrt{7}} \\ 4PD + \sqrt{7}PD & = 4PD - \sqrt{7}PD + 24 - 6\sqrt{7} \\ 2\sqrt{7}PD & = 24 - 6\sqrt{7} \\ PD & = \frac{12}{\sqrt{7}} - 3 = \frac{12\sqrt{7}}{7} - 3 \end{aligned}$
Hence the area is equal to
$\frac{1}{2}\left(\frac{12\sqrt{7}}{7} - 3\right)(4 + \sqrt{7}) = 12 + \frac{69}{14}\sqrt{7}$

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