3. As shown in Figure 3, given that $M$ is a point inside rectangle $A B C D$, $A B=1, B C=2, t=$ $A M \cdot M C+B M \cdot M D$. Then the minimum value of $t$ is $\qquad$
3. 2 .

As shown in Figure 9, draw perpendiculars from point $M$ to $A B$ and $B C$, with the feet of the perpendiculars being $P$ and $Q$ respectively.
Let $A P=a$,
$$
\begin{array}{l}
B P=b, \\
B Q=c, Q C=d .
\end{array}
$$

Then $a+b=1, c+d=2$.
Notice that,
$$
\begin{array}{l}
A M^{2} \cdot M C^{2}=\left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right) \\
=a^{2} b^{2}+a^{2} d^{2}+b^{2} c^{2}+c^{2} d^{2} \\
=(a d+b c)^{2}+(a b-c d)^{2} \geqslant(a d+b c)^{2}
\end{array}
$$

Therefore, $A M \cdot M C \geqslant a d+b c$, with equality holding if and only if $a b=c d$.

Similarly, $B M \cdot M D \geqslant a c+b d$, with equality holding if and only if $a b=c d$.
$$
\begin{array}{l}
\text { Hence } t \geqslant(a d+b c)+(a c+b d) \\
=(a+b)(c+d)=2,
\end{array}
$$

with equality holding if and only if $a b=c d$.
For example, when $a=0.5+0.1 \sqrt{6}, b=0.5-0.1 \sqrt{6}$, $c=0.1, d=1.9$, we have $a b=c d$.
Therefore, the minimum value of $t$ is 2.