Let $S=0$ be the equation of the locus of point $P$ in the plane of square $ABCD$ such that $max(PA, PC)= \frac{1}{\sqrt 2} (PB+PD)$. Find the ratio of the area bounded by the $S=0$ to the given square.
I don't understand how does one get the locus of point $P$ with just this single information. Can somebody please help
Suppose $PA\geq PC$. By Ptolomy inequality for quadrilateral $ABPD$ we have $$PB \cdot s +PD\cdot s \geq PA \cdot s\sqrt{2}$$ with equality iff $P$ is on circumcircle of triangle $ABD$. So $P$ is by hypothetis, on the circumcircle for $ABCD$. So the ratio is $\pi/ 2$.