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Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a convex quadrilateral. Extend line $CD$ past $D$ to meet line $AB$ at $P$ and extend line $CB$ past $B$ to meet line $AD$ at $Q$ . Suppose that line $AC$ bisects $\angle BAD$ . If $AD=\frac{7}{4}$ , $AP=\frac{21}{2}$ , and $AB=\frac{14}{11}$ , compute $AQ$ .

Solution

Solution:

Answer: $\frac{42}{13}$

We prove the more general statement $\frac{1}{AB}+\frac{1}{AP}=\frac{1}{AD}+\frac{1}{AQ}$ , from which the answer easily follows.

Denote $\angle BAC=\angle CAD=\gamma$ , $\angle BCA=\alpha$ , $\angle ACD=\beta$ . Then we have that by the law of sines,
$\frac{AC}{AB}+\frac{AC}{AP}=\frac{\sin (\gamma+\alpha)}{\sin (\alpha)}+\frac{\sin (\gamma-\beta)}{\sin (\beta)}=\frac{\sin (\gamma-\alpha)}{\sin (\alpha)}+\frac{\sin (\gamma+\beta)}{\sin (\beta)}=\frac{AC}{AD}+\frac{AC}{AQ}$
where we have simply used the sine addition formula for the middle step.

Dividing the whole equation by $AC$ gives the desired formula, from which we compute
$AQ=\left(\frac{11}{14}+\frac{2}{21}-\frac{4}{7}\right)^{-1}=\frac{42}{13}.$

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