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This is the 3D (and higher D) version of A claim on partitioning a convex planar region into congruent pieces

  • Is there a 3D convex polyhedral solid that can be cut into 2 mutually congruent non-convex polyhedral solids but not into 2 mutually congruent convex solids? An example with say, 3 replacing 2 will also do.

Remark: I had thought that Noam Elkies's pentagon on the above linked page (it allows partition into 2 non-convex and congruent polygons but not into 2 convex congruent polygons) can be 'lofted' into a pentagonal pyramid - with every horizontal section a scaled down copy of the pentagonal base - that can be cut into 2 congruent non-convex polyhedrons but not into 2 convex and congruent polyhedrons. But it doesn't look obvious that congruence of the 2 pieces can be achieved naturally.

If a polyhedron with the desired property is found, one could try to characterize all polyhedrons that can be cut into 2 congruent non-convex polyhedrons but not into convex congruent polyhedrons. Increasing the number of pieces in the question might make the question more difficult.

Note: As has been noted more recently in the above linked page, even on the plane, I don't know any convex polygon with even number of sides that has the desired property.

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    $\begingroup$ If you just build a prism over Noam Elkies' shape and then cut off some triangular pyramid from the one of the 2 corners with the angle $> 90°$, then the result of putting that thing together with itself is a convex prism with two corners cut off (truncated), still convex. $\endgroup$
    – Wolfgang
    Commented Aug 19 at 12:09
  • $\begingroup$ That's nice! So, there indeed are such polyhedrons. The issue now is to characterize them. Thanks. $\endgroup$
    – Nandakumar R
    Commented Aug 20 at 5:21

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