2023 AMC 10A 第 11 题

Each corner triangle has legs $a,b$ with $a+b=\sqrt3$ (a side of the big square) and $a^2+b^2=2$ (a side of the small square)

$(a+b)^2-(a^2+b^2)$ gives $2ab;$ then $a$ and $b$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0$

Each corner right triangle has legs $a$ and $b.$ A side of the outer square gives $a+b=\sqrt3,$ and a side of the inscribed square gives $a^2+b^2=2.$ Subtract to find the product: $2ab=(a+b)^2-(a^2+b^2)$ $=3-2$ $=1,$ so $ab=\tfrac12.$ Then $a$ and $b$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0,$ namely $\tfrac{\sqrt3\pm1}{2}.$ The ratio of the smaller leg to the larger is $\dfrac{\sqrt3-1}{\sqrt3+1}$ $=\dfrac{(\sqrt3-1)^2}{2}$ $=2-\sqrt3.$ Thus, C is the correct answer.