Tetrahedron splitting/subdivision - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:50:27Z http://mathoverflow.net/feeds/question/28615 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28615/tetrahedron-splitting-subdivision Tetrahedron splitting/subdivision SigTerm 2010-06-18T11:27:18Z 2010-06-19T08:39:09Z <p>Given a regular Tetrahedron <em>A</em> (i.e. each edge of <em>A</em> has same length), is it possible to split <em>A</em> into several smaller regular tetrahedra of equal size? I.e. smaller tetrahedra should completely fill volume of <em>A</em>, and they should not overlap.</p> <p>This can be done in 2D with a triangle and square, and it can be done in 3D with cube (i.e. you can split cube into several smaller cubes of equal size). But I see no way to do same thing in 3D with tetrahedron. </p> <p>If this can be done, how (how smaller tetrahedra should be positioned)?<br> If this cannot be done, is there a proof that this is impossible?</p> <p>P.S. I'm not a mathematician, and this is not a homework, but I'd like to know how/if this can be done.</p> http://mathoverflow.net/questions/28615/tetrahedron-splitting-subdivision/28618#28618 Answer by Anton Petrunin for Tetrahedron splitting/subdivision Anton Petrunin 2010-06-18T11:46:45Z 2010-06-19T08:39:09Z <p><strong>Answer 1:</strong> Look what happens on a face of the big tetrahedron where some edges of small ones come together: you have to make angle 180° out of some dihedral angles of the tetrahedra (which is about 70°) --- that is impossible.</p> <p><strong>Answer 2:</strong> There is the so-called <a href="http://en.wikipedia.org/wiki/Dehn_invariant" rel="nofollow">Dehn invariant</a>. If a polyhedron $X$ is split into a number of polyhedra $Y_1,Y_2,\dots,Y_n$ then the Dehn invariant of $X$ is equal to the sum of the Dehn invariants of $Y_i$.</p> <p>For the regular tetrahedron, the Dehn inveriant is nonzero and proportional to the length of a side. Suppose you could split a regular tetrahedron with side $a$ into a number of tetrahedra with sides $a_1, a_2,\dots, a_n$. Then from the volume you have $$a_1^3+a_2^3+\dots+a_n^3=a^3$$ and from the Dehn invariant you have $$a_1+a_2+\dots+a_n=a.$$ It follows that there is no nontrivial splitting.</p> http://mathoverflow.net/questions/28615/tetrahedron-splitting-subdivision/28620#28620 Answer by Joseph O'Rourke for Tetrahedron splitting/subdivision Joseph O'Rourke 2010-06-18T12:10:47Z 2010-06-18T12:10:47Z <p>The answer is: No. There is a somehwat rambling discussion <a href="http://answers.google.com/answers/threadview/id/497054.html" rel="nofollow">here</a>. Let $B$ be a smaller tetrahedron that is jammed into the apex of $A$. It fills the solid angle there completely. Let $e$ be a base edge of $B$. Then one cannot fill the neighborhood of $e$ by gluing in further regular tetrahedra along it. One way to see this is that the dihedral angle of the tetrahedron is $\delta = \cos^{-1}(1/3) \approx 70.5^\circ$, and the dihedral angle along $e$ to be filled is $\pi - \delta \approx 109.5^\circ$, which cannot be formed from copies of $\delta$.</p> http://mathoverflow.net/questions/28615/tetrahedron-splitting-subdivision/28634#28634 Answer by Henry Segerman for Tetrahedron splitting/subdivision Henry Segerman 2010-06-18T14:29:37Z 2010-06-18T14:29:37Z <p>It is however possible to split a regular tetrahedron into four smaller regular tetrahedra and one regular octahedron, and there are other possibilities based on the <a href="http://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb" rel="nofollow">tetrahedral-octahedral honeycomb</a>.</p> http://mathoverflow.net/questions/28615/tetrahedron-splitting-subdivision/28657#28657 Answer by Mikola for Tetrahedron splitting/subdivision Mikola 2010-06-18T19:11:59Z 2010-06-18T19:11:59Z <p>This question has already been pretty handily answered by Anton, but I figured I'd add another (somewhat ridiculous) proof to his list:</p> <p>Answer 3: Suppose that we had an infinite subdivision of regular tetrahedra. Then we would also have a crystallographic lattice with tetrahedral symmetry, since this structure would have to tile across 3D space. However, the classification of crystallographic lattices rules out this possibility!</p> <p><a href="http://en.wikipedia.org/wiki/Crystallographic_group" rel="nofollow">http://en.wikipedia.org/wiki/Crystallographic_group</a></p>