Let $ABCD$ be a cyclic quadrilateral of area 8. If there exists a point $O$ in the plane of the quadrilateral such that $OA+OB+OC+OD = 8$, prove that $ABCD$ is an isosceles trapezoid.
We have $OA+OB+OC+OD \geq AC+BD \geq 2 \sqrt{AC.BD} \geq 2 \sqrt{2S} = 8$.
The equality happens when $AC = BD$ and $AC$ is perpendicular to $BD$. Hence $ABCD$ is an isosceles trapezoid.
Quite a contrived problem - any such convex quadrilateral will have equal-length, perpendicular diagonals, with only possible positioning for point $O$ at their meeting point ... whence the trivial conclusion - fortunately the other problems in the set were slightly more challenging