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Is there a way to dissect any tetrahedron into a finite number of orthoschemes?

I know that for a tetrahedron which only has acute angles, one can take the center of the inscribed circle and project the center on all the faces and edges and connect it with the vertices to get the orthoschemes. This however does not work when the tetrahedron is allowed to have obtuse angles since the projection of the center of the inscribed circle on the plane containing a face for instance may fall outside of the tetrahedron.

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Yes, this is known (12 is always enough). Interestingly, in higher dimension it is open whether every simplex in $\Bbb R^d$ can be dissected into finitely many orthoschemes (also called path-simplices). This is called Hadwiger's conjecture. See this survey for results and refs to proofs of the conjecture for $d \le 5$.

P.S. In 1993, Tschirpke showed that 12,598,800 orthoschemes suffices in $\Bbb R^5$.

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    $\begingroup$ Yet another "Hadwiger conjecture" to add to the ones about clique minors and covering convex bodies with 2^n smaller homothetic bodies? We need more distinctive nomenclature. $\endgroup$
    – David Eppstein
    Commented May 12, 2010 at 23:34
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    $\begingroup$ There is yet another "Hadwiger's conjecture" saying that every two polytopes with the same volume and Hadwiger invariants are scissor congruent (generalizing Sydler's thm in $\Bbb R^3$ and $\Bbb R^4$ on the volume and Dehn invariant). Still, is nothing compared to en.wikipedia.org/wiki/Erdos_conjecture $\endgroup$
    – Igor Pak
    Commented May 12, 2010 at 23:39
  • $\begingroup$ Hey, Thanks for the response. I read through the paper and unfortunately the references containing the proof that a tetrahedra can be dissected into 12 orthoschemes are in German. Do you know of an English reference that has this material? Thanks $\endgroup$
    – Opt
    Commented Jun 3, 2010 at 15:33
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    $\begingroup$ @Sid. I don't know. I would read Tschirpke's paper springerlink.com/content/l213l5l11jt82187 if you want to get the general idea. If you don't care about 12, but say 100 will work for you, this is a relatively easy exercise which I did some time ago. Hint: start with a "barycentric subdivision" from the inscribed sphere. That gives 24 orthoschemes, and works in many (but not all!) cases. Figure out what goes wrong, cut your simplex into two. Repeat. P.S. You can also use Google Translate which can handle .pdf files. $\endgroup$
    – Igor Pak
    Commented Jun 4, 2010 at 4:22

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