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 P1​P2​, P1​P3​,P1​P4​,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 P1​Pi​​ and P1′​Pi′​​, respectively. If Q2​ and Q3​ were opposite vertices of the octahedron, then the midpoint M of Q2​Q3​​ would be the center of the octahedron. Because M lies on the plane P1​P2​P3​, 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 △P1​Qi​Qj​ gives
s2=(Qi​Qj​)2=(P1​Qi​)2+(P1​Qj​)2
It follows that P1​Q2​=P1​Q3​=P1​Q4​=22​​s, and by symmetry, P1′​Q2′​=P1′​Q3′​=P1′​Q4′​=22​​s. Consequently Qi​ and Qi′​ are opposite vertices of the octahedron. The Pythagorean Theorem on △Q2​P2​P3′​ and △Q3′​P3′​Q2​ gives
(Q2​P3′​)2=(P2​P3′​)2+(Q2​P2​)2=1+(1−22​​s)2 and s2=(Q2​Q3′​)2=(P3′​Q3′​)2+(Q2​P3′​)2=(1−22​​s)2+1+(1−22​​s)2.​
