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Please, please, i need some hint about, tetrahedron

  1. Jan 15, 2008 #1
    Commandino’s Theorem states that

    The four medians of a tetrahedron
    concur in a point that divides each of
    them in the ratio 1:3, the longer
    segment being on the side of the vertex
    of the tetrahedron.

    can someone put links below where about proof of this theorem

    thx so much
  2. jcsd
  3. Jan 17, 2008 #2
    Why not just prove it yourself. If you know anything about dihedral angles, or even general geometry, this should be a piece of cake.
  4. Jan 17, 2008 #3


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    i think i would use vector algebra.
  5. Jan 17, 2008 #4


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    heres another point of view. since linear transformations take medians to medians, and all tetrahedra are equivalent by linear transformations it suffices to prove it for a regular tetrahedron.

    but since the medians of the base triangle meet at a point, and the three medians originating from the base arise by lifting one end of those medians the same distance, it follows at least that those three medians meet in a point directly above the center of the base. hence dropping a median from the top vertex passes through this same point.

    it remains to see where the point is.
  6. Jan 18, 2008 #5
    Question: What condition should a linear transformation hold, in order to preserve ratios of distances? And does that condition will apply to a transformation which is custom-made to map a regular tetrahedron to a non-regular one (a "nice" transformation, since it won't scramble the vertices or something)?

    At the bottom, degenerate case, I can imagine a transformation that always sets the Z coordinate to 0, mapping the regular tetrahedron to the X-Y plane and showing a perspective view of it. Obviously distance ratios are not preserved. Thus there must be a condition that a "nice" transformation satisfies.
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