4. Given that $\alpha-l-\beta$ is a dihedral angle of size $45^{\circ}$, $C$ is a fixed point inside the dihedral angle, and the distances from $C$ to the half-planes $\alpha$ and $\beta$ are $\sqrt{2}$ and 6, respectively. $A$ and $B$ are moving points in the half-planes $\alpha$ and $\beta$, respectively. Then the minimum value of the perimeter of $\triangle A B C$ is $\qquad$
4. $10 \sqrt{2}$.

As shown in Figure 4, construct the symmetric points $P$ and $Q$ of point $C$ with respect to planes $\alpha$ and $\beta$, respectively. It is easy to prove that when $A$ and $B$ are the intersection points of line $P Q$ with planes $\alpha$ and $\beta$, respectively, the perimeter of $\triangle A B C$ is minimized, and this minimum perimeter value is
$$
\begin{array}{l}
|P Q|=\sqrt{(2 \sqrt{2})^{2}+12^{2}-2 \times 2 \sqrt{2} \times 12 \cos \left(180^{\circ}-45^{\circ}\right)} \\
=10 \sqrt{2} .
\end{array}
$$