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.)

let’s set the dimensions of our cube at 1x1x1, and then work with only the upper half of the cube (1x1x height of 1/2) which we will call ‘halfcube’. since each vertex removed takes away 1/8 of the area of the top square face (area of 1), then removing the four vertices of the halfcube reduces the area of the top face to a new square with an area of 1/2.

as an aside, let’s picture a pyramid with the base and the height of the original halfcube, whose volume, according to the formula, would be 1/3 the volume of that halfcube.

back to our problem: the four removed vertices can be arranged to form a pyramid whose height is 1/2 and whose base is the same as the new square top face of our truncated halfcube, that is, an area of 1/2. this pyramid would thus have a volume half that of the above mentioned pyramid. since 1/2 of 1/3 is 1/6, then our truncated halfcube has a volume of 5/6 of the original halfcube. (of course this ratio is the same as it is for removing the 8 vertices of the original FULL cube.)

Nikki had the same answer but she used a whole lot of math that is, to be honest, OVER MY HEAD!

Posted by rixar on 2004-05-26 10:48:14