2023 AMC 12A Problem 9

The outer square has side $\sqrt3$ and the inner square has side $\sqrt2.$ Each triangle is right, with legs $p$ and $q$ along an outer side, so $p+q=\sqrt3,$ and with hypotenuse an inner side, so $p^2+q^2=2.$ Then $(p+q)^2=3$ gives $2pq=1,$ so $p$ and $q$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0,$ namely $\dfrac{\sqrt3\pm 1}{2}.$ The ratio of shorter to longer leg is $$\begin{gathered} \dfrac{\sqrt3-1}{\sqrt3+1}\\ {}=\dfrac{(\sqrt3-1)^2}{2}\\ {}=2-\sqrt3. \end{gathered}$$ Thus, the correct answer is C.