BMO 2010 Shortlist

Let $ABCD$ be a cyclic quadrilateral with $AB=a$, $BC=b$, $CD=c$, $DA=d$, $AC=e$ and $BD=f$. Because the midpoints of the cyclic quadrilateral $ABCD$ form another cyclic quadrilateral, which is a parallelogram, we deduce that this parallelogram is a rectangle and $ABCD$ is orthogonal.

It follows that $$S=\sigma (ABCD)=\frac{ef}{2}$$ and $$a^2+c^2=b^2+d^2$$ Let $R$ and $R_1$ be the circumradii of the cyclic quadrilateral $ABCD$ and the rectangle respectively. We obtain $$4R_1^2=\frac{e^2+f^2}{4}$$ and $$16R^2{S}^2=(ac+bd)(ab+cd)(ad+bc)=ef[(ac(b^2+d^2)+bd(a^2+c^2))]=(ef{)}^2(a^2+c^2),$$ so $$4R^2=a^2+c^2$$ Note that our inequality $R^2\ge 2R_1^2$ is equivalent to $$2(a^2+c^2)\ge e^2+f^2$$ $$a^2+b^2+c^2+d^2-e^2-f^2\ge 0,$$ hence $2(a^2+c^2)\ge (e^2+f^2)$, and we are done. $\square$