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Six Cubes (Posted on 2004-08-08) Difficulty: 4 of 5
Take six identical cubes and place them on the table in such a way that each cube touches each of the other five cubes with some part of its side (touching only along edges or at corners doesn't count).

See The Solution Submitted by SilverKnight    
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Solution how--solution | Comment 2 of 5 |

Start by placing two cubes face-to-face, say arranged horizontally (both in the sense of flat on the table and in the sense as viewed from above in orientation to your eyes). Place a third cube also flat on the table, such that it touches one half face of each of the first two cubes, forming sort of a triangle on the table.  Say the third triangle is "above" the other two (visually, that is--it's actually flat on the table also).

Then construct a similar threesome off to the side.  But turn that threesome 135 degrees, say clockwise, so that visually the lone cube (the one that shares half of one face with each of the other two cubes) is down and to the right (between the 4-o'clock and 5 o'clock positions).  Then, while keeping this whole construct parallel to the table (the centers of the cubes being in a plane parallel to the table). place it above the first layer of cubes, such that the dividing line between the (new) pair of cubes and the (new) single cube cuts across all three lower-layer cubes, and the boundary between the two members of the (new) pair passes above just to the left of the point where the three bottom-layer cubes meet.  The singlet cube at the 4:30 position will be almost directly above the lower right cube on the lower level, but also catch pieces of the other two cubes.  The pair on the higher level (away from the table surface) will individually be mostly above one of the other two cubes on the lower level, but also nick small pieces above each of the other two base cubes.

Edited on August 8, 2004, 8:42 am
  Posted by Charlie on 2004-08-08 08:39:30

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