China Girls' Mathematical Olympiad

$a_{\min }=\frac{1+\sqrt{5}}{2}$.

Write $\phi =\frac{1+\sqrt{5}}{2}$. We may assume that each edge has length $\sqrt{2}$. For any point $P$ inside the square $ABCD$, let $S_1,S_2,S_3,S_4$ denote the area of $△PAB$, $△PBC$, $△PCD$, $△PDA$ respectively; we can also assume that $S_1\ge S_2\ge S_3\ge S_4$.

Let $\lambda =\frac{S_1}{S_2}$, $\mu =\frac{S_2}{S_4}$. If $\lambda ,\mu >\phi$, as $$S_1+S_3=S_2+S_4=1,$$ we have $\frac{S_2}{1-S_2}=\mu$, $S_2=\frac{\mu }{1+\mu }$. So $$S_1=\lambda S_2=\frac{\lambda \mu }{1+\mu }=\frac{\lambda }{1+\frac{1}{\mu }}>\frac{\phi }{1+\frac{1}{\phi }}=\frac{{\phi }^2}{1+\phi }=1,$$ and we reach a contradiction. Hence, $\min \{\lambda ,\mu \}\le \phi$, which implies that $a_{\min }\le \phi$.

On the other hand, for any $a\in (1,\phi )$, we take any $t\in (a,\frac{1+\sqrt{5}}{2})$ such that $b=\frac{t^2}{1+t}>\frac{8}{9}$. Inside the square $ABCD$ we can choose a point $P$ so that $S_1=b$, $S_2=\frac{b}{t}$, $S_3=\frac{b}{t^2}$, $S_4=1-b$. Then we have $$\frac{S_1}{S_2}=\frac{S_2}{S_3}=t\in (a,\frac{1+\sqrt{5}}{2}),$$ $$\frac{S_3}{S_4}=\frac{b}{t^2(1-b)}>\frac{b}{4(1-b)}>2>a.$$ Thus, for any $i,j\in \{1,2,3,4\}$, we have $\frac{S_i}{S_j}\notin [a^{-1},a]$.

Hence $a_{\min }=\phi$.