Source: JBMO Shortlist 2006
Tags: geometry, trapezoid, geometry proposed
Toggle Solutions
Let $ ABCD$ be a trapezoid inscribed in a circle $ \mathcal{C}$ with $ AB\parallel CD$, $ AB=2CD$. Let $ \{Q\}=AD\cap BC$ and let $ P$ be the intersection of tangents to $ \mathcal{C}$ at $ B$ and $ D$. Calculate the area of the quadrilateral $ ABPQ$ in terms of the area of the triangle $ PDQ$.
Denote by $ 2a=AB$ and $ b=AD=CB$. We deduce that $ DC=a$. Denote by $ x=S_{\triangle PQD}$ and by $ h$ the altitude the triangle $ QAD$.
$ \angle CBP=\angle QDP$, $ CB=DQ$ and $ PD=PB$ $ \Rightarrow S_{\triangle QDP}=S_{\triangle PCB}$ $ \Rightarrow S_{\triangle QPA}=S_{\triangle QPB} \Rightarrow ABPQ$ is a trapezoid.
Denote by $ R=DP \cap AB \Rightarrow \triangle RAD \sim \triangle ADC \Rightarrow AD^2=RA \cdot DC \Rightarrow RA=\frac {b^2} {a}$.But $ PQ=RA \Rightarrow PQ=\frac {b^2} {a}$
$ S_{\triangle PQA}=2x=\frac {h \cdot PQ} {2} \Rightarrow h=\frac {4x} {PQ}=\frac {4ax} {b^2}$
$ S_{ABPQ}=S_{ABQ}+S_{BPQ}=\frac {h \cdot 2a} {2}+2x=\frac {4ax \cdot 2a} {2 \cdot b^2}+2x=\frac {4a^2x} {b^2}+2x$
I think there is a mistake, you can't take altitude of ADQ because these three points are colinear.