8. The answer is (D). Let $L$ be the point on the floor where the thief "lands", $D$ the point where the diamond is located; let $D_{E}, D_{W}, D_{N}, D_{S}$ be the symmetrical points of $D$ with respect to the east, west, north, and south walls $(E, W, N, S$ in the figure). Imagine that the thief touches the wire (to cut it) at a point $P$ on the wall $E$, then goes to pick up the diamond at $D$: the path he takes is at least $L P + P D$, which by symmetry is equal to $L P + P D_{E}$; therefore, assuming the thief cuts the wire at a point on the wall $E$, the shortest possible path, as a consequence of the triangle inequality, is the one where $P$ is aligned with $L$ and $D_{E}$ (this point is indicated with $P_{E}$ in the figure), and is $\sqrt{4^{2}+1}=\sqrt{17}$ meters long.

Repeating the same reasoning for the other three walls (and noting, to save calculations, that $L D_{E}=L D_{W}, L D_{S}=L D_{N}$) reduces the problem to comparing the lengths of $L D_{E}$ and $L D_{N}$; it turns out that the minimum distance to be traveled is $\sqrt{2^{2}+3^{2}}=\sqrt{13}$ meters.
8. The tiny but precious Dodecahedral Diamond is located 2 meters from the south wall and 3 meters from the west wall of a rectangular room whose north and south walls are 4 meters long and the east and west walls are 3 meters long. A thief drops from the ceiling inside the room and touches the floor one meter from the south wall and one meter from the west wall. However, he realizes that he must immediately deactivate the alarm system by cutting at least one point of a wire that runs at a constant height along the four perimeter walls of the room. How long is the shortest path he must take to reach a point on one of the walls first, and then the Dodecahedral Diamond?
(A) $3+\sqrt{2}$
(B) $2+\sqrt{5}$
(C) $\sqrt{17}$
(D) $\sqrt{13}$
(E) $2 \sqrt{2}$.