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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 can further ask how 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.

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.

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.

One can further ask how 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.

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 can further ask how 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.

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: It appears 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 slanted pentagonal pyramid that can be cut into 2 congruent non-convex polyhedrons but not into 2 convex and congruent polyhedrons. However, one could still ask if there are other kinds of convex polyhedrons with the desired partition property.

One can further ask how 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.

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.

One can further ask how 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.