Let the length be L and the width W. Each, we are told, is a one-digit number.
The volume of the whole block is W^2 * L.
The volume cut off by the first cut is that of a pyramid whose base is half the square face and whose height is L, and is thus L*W^2 / 6.
The volume cut off by the second cut would be the same as the first cut except for the portion occupied by both pyramids.
The portion occupied by both pyramids is a tetrahedron whose vertices are D, D', M, M', where M is the center of rectangle DAA'D', and M' is the center of rectangle DCC'D'. Considered as a pyramid, the base can be considered as triangle DMA', with area L*W/4 and height W/2, for a volume of L*W^2/24.
The volume of the largest remaining piece is therefore, by inclusion/exclusion:
W^2 * L - 2 * L*W^2 / 6 + L*W^2/24
Tabulated for single-digit dimensions, this comes out to:
L \ W 1 2 3 4 5 6 7 8 9
1 0.708 2.833 6.375 11.333 17.708 25.500 34.708 45.333 57.375
2 1.417 5.667 12.750 22.667 35.417 51.000 69.417 90.667 114.750
3 2.125 8.500 19.125 34.000 53.125 76.500 104.125 136.000 172.125
4 2.833 11.333 25.500 45.333 70.833 102.000 138.833 181.333 229.500
5 3.542 14.167 31.875 56.667 88.542 127.500 173.542 226.667 286.875
6 4.250 17.000 38.250 68.000 106.250 153.000 208.250 272.000 344.250
7 4.958 19.833 44.625 79.333 123.958 178.500 242.958 317.333 401.625
8 5.667 22.667 51.000 90.667 141.667 204.000 277.667 362.667 459.000
9 6.375 25.500 57.375 102.000 159.375 229.500 312.375 408.000 516.375
Only L=3, W=4 shows the volume as a 2-digit integer whose digits represent the dimensions of the original solid. The remaining piece therefore has volume **34 cm^3**.
From Enigma No. 1456, by Susan Denham, New Scientist 18 August 2007. |