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2012 AMC 12B Problem 19

Problem 19 of 25HarderGeometry

A unit cube has vertices P1,P_1, P2,P_2, P3,P_3, P4,P_4, P1,P_1', P2,P_2', P3,P_3', and P4.P_4'. Vertices P2,P_2, P3,P_3, and P4P_4 are adjacent to P1,P_1, and for 1i4,1 \le i \le 4, vertices PiP_i and PiP_i' are opposite to each other. A regular octahedron has one vertex in each of the segments P1P2,P_1P_2, P1P3,P_1P_3, P1P4,P_1P_4, P1P2,P_1'P_2', P1P3,P_1'P_3', and P1P4.P_1'P_4'. What is the octahedron’s side length?

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Solution

Place P1P_1 at the origin with edges along the axes, and let each of the three octahedron vertices near P1P_1 be a distance tt from P1;P_1; by symmetry the three near P1P_1' are also a distance tt from P1.P_1'. Two vertices sharing P1,P_1, such as (t,0,0)(t,0,0) and (0,t,0),(0,t,0), are a distance t2t\sqrt2 apart. A vertex near P1,P_1, say (t,0,0),(t,0,0), and the appropriate vertex near P1,P_1', say (1,1t,1),(1,1-t,1), must be the same distance apart. Setting the two squared side lengths equal and using the cube’s unit edges yields t=34,t=\tfrac34, so the side length is t2=324.t\sqrt2=\dfrac{3\sqrt2}{4}. Thus, the correct answer is A.

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Concepts: cube geometry · Pythagorean Theorem · symmetry

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.