23 Nov 1998
Divide the Soma pieces into two categories - the planar pieces and the non-planar pieces. All three of the non-planar pieces have a size of 4, while three of the four planar pieces are of size 4 and one is of size 3. I'll refer to the 3 x 4 slabs on the right and left sides of the arch the "sides" and the three remaining cubes running along the roof of the arch I'll call the ceiling.
Since none of the non-planar pieces can lie entirely within one of the sides, each of them must account for exacty one of the 3 cubes in the ceiling. This implies that none of the planar pieces can span the arch and intersect both sides. Since all the non-planar pieces are of diameter of 2, it follows more generally that each of the seven Soma pieces intersects one and only one of the two sides of the arch.
Now there isn't room for all three of of the non-planar pieces to intersect the same side so one of the sides of the arch must intersect with exactly two of them. This accounts for 6 of the 12 cubes on that side. The 6 remaining cubes on that side must be exactly filled by some combination of the four planar pieces, three with size 4 and one with size 3. This is clearly impossible. Therefore, there is no way to build this figure with the Soma cubes.
- Proved and submitted by Richard Sullivan
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