Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 25I / b / Solution

Let $A$ and $C$ be opposite interior angles of a convex quadrilateral with side lengths $a,b,c,d$, and let $s=(a+b+c+d)/2$. The Bretschneider formula gives its area $K$ asThe fixed side lengths fix the first term, while the second term is nonpositive. Hencewith equality exactly when $A+C=\pi$. Opposite angles are supplementary exactly when the quadrilateral is a cyclic quadrilateral. The given $Q$ is cyclic, so it attains this upper bound andA crossed or concave competitor can be uncrossed or reflected across a diagonal without changing its side lengths and without decreasing its unsigned area, so the same bound applies. Apart from degenerate coincidences, equality holds precisely when $Q^{\prime }$ is also cyclic with the stated cyclic ordering of its side lengths.

Solved by gpt-5.6-sol high.