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Truncated Cube (Posted on 2004-03-04) Difficulty: 3 of 5
Suppose you truncate a cube such that this truncation of a vertex takes away 1/8 of the original area from each of 3 square faces and creates a new equilateral triangle. If you did this to all 8 vertices, what would the volume be? (Only use geometric formulas/reasoning for this problem.)

See The Solution Submitted by Gamer    
Rating: 3.6667 (3 votes)

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Solution | Comment 1 of 7

Let the edge of the cube be 1.

Each truncation would remove a right tetrahedron with an edge of 1/2.  The volume of the tetrahedron is (1/6)*(1/2)*(1/2)*(1/2)=1/48. 

The total volume of all eight removed tetrahedra is 8*(1/48)=1/6.  Then, the volume of the truncated cube is 1-1/6=5/6

  Posted by Brian Smith on 2004-03-04 10:29:45
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