You have a 3x3x3 cube. You bore a circular hole, of diameter 1, from the center of the top face to the center of the bottom face. You bore a similar hole from the left face center to the right face center, and another hole from front center to back center. How much material was removed?

It seems to me that the answer is rather straight-forward (except for the mat involved in the final step)

• Take the volume of the cube (9)
• Subtract the volume of the first cylander [9-3¥ð(¨ö©÷)=9-¨ú¥ð]
• Subtract the volume of the second cylander [9-3¥ð/2]
• Add back the volume of the mouhefanggai (2-cylander Steinmetz solid) [9-3¥ð/2+16/3]
• Subtract the volume of the third cylander [9-9¥ð/4+16/3]
• Add back the volumes of the three-cylander Steinmetz solid -- and the pieces that are the intersections of only two cylanders -- (not Charlies' "void," which also includes volumes in only one cylander)

OK, after realizing that it is not just the three-cylander Steinmetz solid that needs to be added back, and adding the italicized portion above (although the phrase not Charlie's "void" was original), I  have to admit Charlie is not making things "overly complicated," but just approaching a severely complicated problem from a differen direction. 

Sorry, Charlie (Starkist doesn't want tuna with "good taste"; Starkist wants tuna that tastes good.) ;-)

Posted by TomM on 2005-03-08 02:54:36