Problem:
A unit cube has vertices P1,P2,P3,P4,P1′,P2′,P3′, and P4′. Vertices P2,P3, and P4 are adjacent to P1, and for 1≤i≤4, vertices Pi and Pi′ are opposite to each other. A regular octahedron has one vertex in each of the segments P1P2, P1P3,P1P4,P1′P2′,P1′P3′, and P1′P4′. What is the octahedron's side length?
Answer Choices:
A. 432
B. 1676
C. 25
D. 323
E. 26 Solution:
Let s be the length of the octahedron's side, and let Qi and Qi′ be the vertices of the octahedron on P1Pi and P1′Pi′, respectively. If Q2 and Q3 were opposite vertices of the octahedron, then the midpoint M of Q2Q3 would be the center of the octahedron. Because M lies on the plane P1P2P3, the vertex of the octahedron opposite Q4 would be outside the cube. Therefore Q2,Q3, and Q4 are all adjacent vertices of the octahedron, and by symmetry so are Q2′, Q3′, and Q4′. For 2≤i<j≤4, the Pythagorean Theorem applied to △P1QiQj gives
s2=(QiQj)2=(P1Qi)2+(P1Qj)2
It follows that P1Q2=P1Q3=P1Q4=22s, and by symmetry, P1′Q2′=P1′Q3′=P1′Q4′=22s. Consequently Qi and Qi′ are opposite vertices of the octahedron. The Pythagorean Theorem on △Q2P2P3′ and △Q3′P3′Q2 gives
(Q2P3′)2=(P2P3′)2+(Q2P2)2=1+(1−22s)2 and s2=(Q2Q3′)2=(P3′Q3′)2+(Q2P3′)2=(1−22s)2+1+(1−22s)2.
