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This is driven more by curiosity than by research, but nevertheless may be of some interest.

Start with a regular tetrahedron $T$ with corners $(a,b,c,d)$, and let $x$ be its incenter—the center of the largest inscribed sphere. Partition $T$ into four tetrahedra, with corners at $$(a,b,c,x), \; (a,b,d,x), \; (a,c,d,x), \; (b,c,d,x) \;.$$ And now iterate the process: Find the incenters of those four tetrahedra, partition each etc. Connecting the incenters forms a tree, whose convex hull is a regular tetrahedron:


          TetraReg
I was hoping that the same process for an irregular tetrahedron would lead to a similar tetrahedral hull, but it seems more complicated:
          Tetrar16
The question is:

Q. To what shape does the hull of the incenter tree approach as the number of iterations goes to $\infty$?

Perhaps it still approaches a tetrahedron, but just much more slowly? The above shows $8$ iterations.

(Incidentally, replacing "incenter" by "centroid" appears to lead to polyhedra that approach filling out the entire interior of the original $T$—less interesting behavior.)

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