IRL_ABooklet_2023

Let $a=|AB|$, $b=|BC|$, $c=|CD|$, $d=|DA|$ be the side lengths of $Q$, and $f=|AC|$, $g=|BD|$ the lengths of its diagonals. It's well-known and easy to prove that $$f^2=\frac{ab(c^2+d^2)+cd(a^2+b^2)}{ab+cd},$$ and $$g^2=\frac{ad(b^2+c^2)+bc(a^2+d^2)}{ad+bc}.$$ For instance, to obtain the formula for $g$, apply the Cosine Rule in each of the triangles $ABD$ and $BCD$. Then $$\cos A=\frac{a^2+d^2-g^2}{2ad},\cos C=\frac{b^2+c^2-g^2}{2bc}.$$ But, since $Q$ is cyclic, $\cos A=-\cos C$, whence $$\begin{array}{rl}0& =2abcd(\cos A+\cos C)\\ & =bc(a^2+d^2-g^2)+ad(b^2+c^2-g^2)\\ & =bc(a^2+d^2)+ad(b^2+c^2)-(ad+bc)g^2.\end{array}$$ Therefore $$g^2=\frac{ad(b^2+c^2)+bc(a^2+d^2)}{ad+bc},$$ i.e., $g^2=\lambda (a^2+d^2)+\mu (b^2+c^2)$, where $$\lambda =\frac{bc}{ad+bc},\mu =\frac{ad}{ad+bc},$$ so that $\lambda ,\mu$ are positive numbers that sum to 1. Thus $g^2$ is a convex sum of $a^2+d^2$ and $b^2+c^2$, and so $g^2\le \max (a^2+d^2,b^2+c^2)\le 2\max (a^2,d^2,b^2,c^2)$, whence $$g\le \sqrt{2}\max (a,b,c,d).$$

Suppose $a=|AB|$ is the longest side, and the diagonal $g$ is equal to $\sqrt{2}a$. Then $$2a^2=\lambda (a^2-d^2)+\mu (2a^2-b^2-c^2)=0.$$ But $0<\lambda ,\mu$ and $a^2\ge \max (b^2,c^2,d^2)$. Hence $d=b=c=a$ and $Q$ is a rhombus. Because $Q$ is cyclic, it must then be a square.