Skip to main content
http -> https (the question was bumped anyway)
Source Link
Martin Sleziak
  • 4.7k
  • 4
  • 35
  • 40

In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problemHilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\ $ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).

In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\ $ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).

In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\ $ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).

Post Made Community Wiki
Source Link
Igor Pak
  • 17k
  • 2
  • 61
  • 123

In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\ $ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).