"Trevor G Quayle" <Tin### [at] hotmailcom> wrote:
> Would a shape with a square top and circular bottom work (similar to 3 which is
> a line top and round bottom).
>
> I don't know what the exact volume is but it would seem that it would be the
> average of a cylinder and a cube: (6+8)/2=7

The problem here is that the targeted volume is (pi/3)*7, not 1*7, and the
volume of the cylinder is (pi/3)*6, not 1*6, while the cube's volume is plain
boring 1*8.

So a gradual cross-section morph from square to circle in a way that the
cross-section area increases linearly (similar to #3) would indeed give a
volume equal to the average of a cylinder's and a cube's volume, but that
wouldn't fit the bill.


Of course, a rectangular box of size 2*2*(pi*7/12), skewed by 2-(pi*7/12) to
"fill" a 2*2*2 cube, would do the job, but something tells me that it would not
be quite in line with the spirit of the puzzle ;)

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