7. Yet another trapezoid $A B C D$ has $A D$ parallel to $B C$. $A C$ and $B D$ intersect at $P$. If $[A D P] /[B C P]=1 / 2$, find $[A D P] /[A B C D]$. (Here the notation $\left[P_{1} \cdots P_{n}\right]$ denotes the area of the polygon $P_{1} \cdots P_{n}$.)
Solution: $3-2 \sqrt{2}$
A homothety (scaling) about $P$ takes triangle $A D P$ into $B C P$, since $A D, B C$ are parallel and $A, P, C ; B, P, D$ are collinear. The ratio of homothety is thus $\sqrt{2}$. It follows that, if we rescale to put $[A D P]=1$, then $[A B P]=[C D P]=\sqrt{2}$, just by the ratios of lengths of bases. So $[A B C D]=3+2 \sqrt{2}$, so $[A D P] /[A B C D]=1 /(3+2 \sqrt{2})$. Simplifying this, we get $3-2 \sqrt{2}$.