Maths Olympiad Prep

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Problem 1457

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Geometry Difficulty 5.9 Multiple choice Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO · Italy

The tiny, but extremely 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 whose east and west walls are 3 meters long. A thief lowers himself from the ceiling into the room and touches the floor at a point 1 meter from the south wall and 1 meter from the west wall. However, he realizes that he must immediately disable the alarm system by cutting, in at least one point, a wire that runs at a constant height from the floor along the four perimeter walls of the room. How many meters long is the shortest path he must take to first reach any point on one of the walls, and then reach the Dodecahedral Diamond?

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Official solution

Solution:

The answer is (D). Let LL be the point on the floor where the thief "lands", DD the point where the diamond is located; let DE,DW,DN,DSD_{E}, D_{W}, D_{N}, D_{S} be the reflections of DD with respect to the east, west, north, south walls (E,W,N,S(E, W, N, S in the figure). Let us imagine that the thief touches the wire (in order to cut it) at a point PP on wall EE, and then goes to fetch the diamond at DD: the path he takes has length (at least) LP+PDL P+P D, which by symmetry is equal to LP+PDEL P+P D_{E}; hence, assuming that the thief cuts the wire at a point on wall EE, the shortest possible path, as a consequence of the triangle inequality, is the one in which PP is aligned with LL and DED_{E} (this point is denoted by PEP_{E} in the figure), and it has length 42+1=17\sqrt{4^{2}+1}=\sqrt{17} meters.

Repeating the same reasoning for the other three walls (and noting, in order to save calculations, that LDE=LDW,LDS=LDNL D_{E}=L D_{W}, L D_{S}=L D_{N}) we reduce to comparing the lengths of LDEL D_{E} and LDNL D_{N}; it follows that the minimum distance to be traveled is 22+32=13\sqrt{2^{2}+3^{2}}=\sqrt{13} meters.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty, ordering) added by this project.