3. Find the smallest constant $a(a>1)$, such that for any point $P$ inside the square $ABCD$, there are always at least two of the triangles $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$, whose area ratio belongs to the interval $\left[a^{-1}, a\right]$.
(Li Weiguo)
3. $a_{\min }=\frac{1+\sqrt{5}}{2}$.

First, prove: $a_{\text {min }} \leqslant \frac{1+\sqrt{5}}{2}$.
Let $\varphi=\frac{1+\sqrt{5}}{2}$. Assume the side length of the square is $\sqrt{2}$.
As shown in Figure 2, for a point $P$ inside the square $A B C D$, let $S_{1} 、 S_{2} 、 S_{3} 、 S_{4}$ represent the areas of $\triangle P A B 、 \triangle P B C$ 、 $\triangle P C D 、 \triangle P D A$ respectively. Assume
$$
\begin{array}{l}
S_{1} \geqslant S_{2} \geqslant S_{4} \geqslant S_{3} . \\
\text { Let } \lambda=\frac{S_{1}}{S_{2}}, \mu=\frac{S_{2}}{S_{4}} .
\end{array}
$$

If $\lambda, \mu>\varphi$, from $S_{1}+S_{3}=S_{2}+S_{4}=1$, we get $\frac{S_{2}}{1-S_{2}}=\mu$, thus, $S_{2}=\frac{\mu}{1+\mu}$.
$$
\begin{array}{l}
\text { Hence } S_{1}=\lambda S_{2}=\frac{\lambda \mu}{1+\mu} \\
=\frac{\lambda}{1+\frac{1}{\mu}}>\frac{\varphi}{1+\frac{1}{\varphi}}=\frac{\varphi^{2}}{1+\varphi}=1,
\end{array}
$$

which is a contradiction.
Therefore, $\min \{\lambda, \mu\} \leqslant \varphi$, indicating $a_{\text {min }} \leqslant \varphi$.
Conversely, for any $a \in(1, \varphi)$, choose $t \in\left(a, \frac{1+\sqrt{5}}{2}\right)$ such that $b=\frac{t^{2}}{1+t}>\frac{8}{9}$.
In the square $A B C D$, take a point $P$ such that
$$
\begin{array}{l}
S_{1}=b, S_{2}=\frac{b}{t}, S_{3}=\frac{b}{t^{2}}, S_{4}=1-b . \\
\text { Then } \frac{S_{1}}{S_{2}}=\frac{S_{2}}{S_{3}}=t \in\left(a, \frac{1+\sqrt{5}}{2}\right), \\
\frac{S_{3}}{S_{4}}=\frac{b}{t^{2}(1-b)}>\frac{b}{4(1-b)}>2>a .
\end{array}
$$

This shows that for any $i 、 j \in\{1,2,3,4\}$, $\frac{S_{i}}{S_{j}} \notin\left[a^{-1}, a\right]$. This indicates $a_{\text {min }}=\varphi$.