Hong Kong Preliminary Selection Contest

Let $r$ be the radius of the circle, and $E$ and $F$ be the points where the circle touches $AD$ and $BC$ respectively. Then $PE=PF=r$ and they are the heights of $△PAD$ and $△PBC$ from $P$. Let $h$ be the height of the trapezium. By considering the areas of $△PAD$ and $△PBC$, we get $\frac{PA\cdot h}{2}=\frac{15r}{2}$ and $\frac{PB\cdot h}{2}=\frac{20r}{2}$, and hence $\frac{PA}{PB}=\frac{3}{4}$. Together with $AB=42$, we have $PA=18$ and $PB=24$. It follows that $PA\times PB=432$.