# Dissecting a tetrahedron into orthoschemes

Hey,

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.

Thanks

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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 $\Bbb R^5$.
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 –  Igor Pak May 12 '10 at 23:39