Three cubes of volume 1, 8, and 64 are glued together.
What is the smallest possible surface area of the resulting configuration?
Problem taken from UNL Math Day
Individually, the surface areas are 1, 24, and 96, for a total of 121. Gluing the 1 and 8 cubes together reduces the surface by 2. Gluing the 1 and 64 cube together reduces the surface by 2. Gluing the 8 and 64 cube together reduces the area by 8. If we can mange a configuration where all three are glued together, it will reduce the area by ten for a new total of 111.
One such configuration is to set the two smaller cubes on top of the large one, abtting one another and glue them together that way. Then glue this configuration in place on the large cube.

Posted by TomM
on 20021224 23:48:54 