6. In tetrahedron $ABCD$, $AB=2\sqrt{3}$, $BC=\sqrt{3}$, $AC=AD=BD=CD=3$, $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively. A beetle wants to travel from point $M$ to point $N$ along the surface of the tetrahedron. To minimize the distance traveled, which edge must it cross?
(A) $AC$
(B) $BC$
(C) $AD$
(D) $BD$
6. A.

From the side length conditions, it is easy to see that $\triangle A B C$ is a right triangle, with $\angle A C B=90^{\circ}, \angle A B C=60^{\circ}, \angle B A C=30^{\circ}$.

Let the shortest path length from point $M$, crossing edge $A C$ to reach point $M$ be denoted as $(A C)$, and similarly for other cases.
(1) First, compare $(A C)$ with $(B C)$.
As shown in Figure 3, rotate $\triangle D A C$ and $\triangle D B C$ around $A C$ and $B C$ respectively, and flatten them onto the plane where $\triangle A B C$ lies (in all figures, the notations $N_{k}$, $C_{k}$, $D_{k} (k=1,2, \cdots)$ represent the new positions of points $N$, $C$, $D$ in the flattened figures).

In $\triangle C N_{1} M$ and $\triangle C N_{2} M$, two pairs of sides are equal $\left(C N_{1}=C N_{2}, C M=C M\right)$, and $\angle M C N_{2} > \angle M C N_{1}$ (since $\angle M C N_{1}=\angle M C A+\angle A C D_{1}=30^{\circ}+60^{\circ}=90^{\circ}, \angle M C B=60^{\circ}, \angle B C D_{2}>60^{\circ}$, then $\left.\angle M C N_{2}>120^{\circ}\right)$, hence $M N_{1} < M N_{2}$, i.e., $(A C) < (B C)$.

Similarly, we can compare $(A C)$ with $(B D)$. In $\triangle C N_{1} M$ and $\triangle D N_{2} M$, two pairs of sides are equal $\left(C N_{1}=D N_{2}, C M=C M\right)$, and $\angle M C N_{1} < \angle M D N_{2}$ (since $\angle M C N_{1}=90^{\circ}, \angle M D N_{2} > 120^{\circ}$), hence $M N_{1} < M N_{2}$, i.e., $(A C) < (B D)$.

According to the above comparisons, $(A C)$ is the shortest.