1. Jan 15, 2008

### wowolala

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.

thx so much

2. Jan 17, 2008

### rbzima

Why not just prove it yourself. If you know anything about dihedral angles, or even general geometry, this should be a piece of cake.

3. Jan 17, 2008

### mathwonk

i think i would use vector algebra.

4. Jan 17, 2008

### mathwonk

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.

5. Jan 18, 2008

### dodo

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.