bonanova Posted January 2, 2009 Report Share Posted January 2, 2009 If rearrangement of the pieces between cuts in not permitted, a cube can be cut into 27 cubes by slicing twice in each of three perpendicular planes. By permitting rearrangement of the pieces, [slicing and stacking], can this be done with fewer than 6 cuts? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 2, 2009 Report Share Posted January 2, 2009 yes Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 2, 2009 Report Share Posted January 2, 2009 I worked on this for 1/2 hour...I discovered that minimum 6 slices are required, because the most middle cube always needs a new cut to be separated from its each adjacent cube. Since it has 6 sides and 6 adjacent cubes, 6 cut is mandatory. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted January 3, 2009 Author Report Share Posted January 3, 2009 I worked on this for 1/2 hour...I discovered that minimum 6 slices are required, because the most middle cube always needs a new cut to be separated from its each adjacent cube. Since it has 6 sides and 6 adjacent cubes, 6 cut is mandatory. Nice proof. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 3, 2009 Report Share Posted January 3, 2009 reminder Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted January 3, 2009 Author Report Share Posted January 3, 2009 Oops. Thanks LIS Thread locked. Quote Link to comment Share on other sites More sharing options...
Question
bonanova
If rearrangement of the pieces between cuts in not
permitted, a cube can be cut into 27 cubes by slicing
twice in each of three perpendicular planes.
By permitting rearrangement of the pieces, [slicing
and stacking], can this be done with fewer than 6 cuts?
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