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- Newsgroups: sci.math
- Path: sparky!uunet!stanford.edu!CSD-NewsHost.Stanford.EDU!SAIL.Stanford.EDU!rivin
- From: rivin@SAIL.Stanford.EDU (Igor Rivin)
- Subject: Re: How's your geometric intuition?
- Message-ID: <1993Jan2.203339.25416@CSD-NewsHost.Stanford.EDU>
- Sender: news@CSD-NewsHost.Stanford.EDU
- Organization: Computer Science Department, Stanford University.
- References: <1992Dec31.222328.22744@samba.oit.unc.edu>
- Date: Sat, 2 Jan 1993 20:33:39 GMT
- Lines: 32
-
- In article <1992Dec31.222328.22744@samba.oit.unc.edu> Jim.Buddenhagen@launchpad.unc.edu (Jim Buddenhagen) writes:
- >Start with a regular tetrahedron circumscribed by a sphere. Subdivide
- >each triangle into 4 triangles by connecting edge midpoints, and project
- >to the sphere (radially), to obtain a new polyhedron with all vertices
- >on the sphere and with 16 triangular faces.
- >
- >Repeat this procedure with the new polyhedron to get a third polyhedron.
- >
- >True or false: this third polyhedron is convex.
- >
- >(I found the answer surprising, what about you?)
- >--
-
- A simpler example of this phenomenon can be gotten by taking a regular
- octahedron inscribed in the sphere and stellating the faces (that is,
- adding a vertex in the middle of each face, connected to the vertices
- of each face), and then pushing the new vertices out to the sphere.
- This will not be convex, no matter where you put the new vertices.
- What's more, there is no convex polyhedron inscribed in the sphere
- combinatorially equivalent to the stellated icosahedron (which only
- has 24 faces, but is not the smallest example).
-
- These kinds of questions go back a long way (to Descartes), and seem
- to be finally well understood, through some work of mine on hyperbolic
- geometry; see the research announcement in the most recent (Oct 92)
- Bulletin of the AMS. I can send you the actual preprints/reprints, if
- you are interested.
-
-
- Igor Rivin
-
-
-