Enigma 1456 — Cutting Edge

by Susan Denham

I started with a small rectangular block with one end a square, ABCD, and with a correspondingly labelled square at the other end, A' B' C' D' . I made two cuts through the block, one through the plane ACD'  and the other through the plane A' C' D. This cut the block into four pieces, and I discarded the smallest three of these.

The volume of the remaining piece is a two-figure number of cubic centimetres. By coincidence, in that number each of the two digits was the length of one of the sides of the original rectangular block.

What is the volume of this remaining piece?

The black lines show the block before we start to cut.

The first cut is shown in red: ACD' 

This will remove a tetrahedron, perhaps best thought of as a pyramid with a triangular base ACD with its apex at D' . (The volume of a pyramid is 1/3 area of the base times the height.)


The second cut is shown in purple: A' C' D

This will remove a tetrahedron, perhaps best thought of as a pyramid with a triangular base A' C' D'  with its apex at D, except that part of that tetrahedron was already removed by the first cut, so we need to find the volume of the bit that was “removed twice”.
This bit that was “removed twice” has the vertices D, X, D'  and Z. It is composed of two pyramids with their bases (XYZ) stuck together. So the volume of the remaining chunk is given by:
          volume of the original block
       −  volume of the pyramid formed by the first cut
       −  volume of the pyramid formed by the second cut
       +  volume of the bit that was “removed twice”
Let the side of the square ABCD be s and the length A A'  be l.
          volume of the original block = s2 l
          volume of the pyramid ACD'   =  volume of the pyramid A' C' D   =   s2 l / 6

The volume of half of the bit that was “removed twice” is that of a “half-sized” pyramid, i.e. 1/8 of the volume of the first cut pyramid.
       =    s2 l / 48

The volume of the remaining chunk is thus given by:
       =  s2 l   −   s2 l / 6   −   s2 l / 6   +   s2 l / 24    =    17 s2 l / 24
         which has therefore to be a multiple of 17 which is less that 100, i.e. 17, 34, 51, 68 or 85.
Furthermore. s2 l has to a multiple of 24, of which the prime factors are 3×2×2×2.
Only 34 and 68 offer digits with the right factors, and 62×8 > 100, so the volume is 34 cm2, which fits with s = 4   and   l = 3.