3. In $\triangle A B C$, $A B=2 A C$, and $S_{\triangle A B C}=1$. Then the minimum value of $B C$ is

In $\triangle A B C$, $A B=2 A C$, and $S_{\triangle A B C}=1$. Then the minimum value of $B C$ is
3. $\sqrt{3}$.

Let $A C=b$.
From $S_{\triangle A B C}=1 \Rightarrow b^{2} \sin A=1 \Rightarrow b^{2}=\frac{1}{\sin A}$.
By the cosine rule, we get
$$
\begin{array}{l}
B C^{2}=5 b^{2}-4 b^{2} \cos A \\
=\frac{5-4 \cos A}{\sin A}=\frac{1+9 \tan ^{2} \frac{A}{2}}{2 \tan \frac{A}{2}}
\end{array}
$$
$\geqslant 3$.
Therefore, $B C \geqslant \sqrt{3}$, with equality holding if and only if $\tan \frac{A}{2}=\frac{1}{3}$.